10. 1.29 dag of sodium chloride is reacted with 25.0 dag of silver nitrate10. Balanced Chemical Equation: Reaction Type:At completion of reactions: Formula of Reactant A: Grams of Reactant A: Formula of Reactant B:Grams of Reactant B: Formula of Product C:Grams of Product C: Formula of Product D:Grams of Product D:

Answers

Answer 1

So,

Let's first convert some units:

[tex]\begin{gathered} 1.29dag\cdot\frac{10g}{1dag}=12.9g \\ \\ 25.0\text{dag}\cdot\frac{10g}{1\text{dag}}=250g \end{gathered}[/tex]

So,

There's a reaction between 12.9g of sodium chloride and 250g of silver nitrate.

The reaction that occurs here, is the next one:

[tex]NaCl+AgNO_3\to AgCl+NaNO_3[/tex]

Sodium chloride reacts with silver nitrate to form silver chloride and sodium nitrate.

This reaction is a displacement reaction.

Now, let's complete the information:

Formula of Reactant A: NaCl

Grams of Reactant A: 12.9g

Formula of Reactant B: AgNO3

Grams of Reactant B: 250g

Formula of Product C: AgCl

Grams of Product C: ?

Formula of Product D: NaNO3

Grams of Product D: ?

We need to find the amount of grams of the formed products.

So, let's first convert the grams of each reactant in moles. Remember that the number of moles (n) can be found if we divide the given mass of the given compound by its molar mass. So,

[tex]\begin{gathered} n_{NaCl}=\frac{12.9gNaCl}{\frac{58.5gNaCl}{\text{mol}}}=0.22\text{molesNaCl} \\ _{} \\ n_{AgNO_3}=\frac{250gAgNO_3}{\frac{169.87gAgNO_3}{mol}}=1.47molesAgNO_3 \end{gathered}[/tex]

The limiting reactant in this reaction is NaCl. So, we're going to use the reaction's stoichiometry to find the amount of the products as follows:

[tex]0.22mol\text{NaCl}\cdot\frac{1\text{molAgCl}}{1\text{molNaCl}}\cdot\frac{143.32\text{gAgCl}}{1\text{molAgCl}}=31.53\text{gAgCl}[/tex]

Thus, there are 31.53g of AgCl formed.

Finally, let's find the amount of grams of product D (NaNO3):

[tex]0.22mol\text{NaCl}\cdot\frac{1\text{molNaNO}3}{1\text{molNaCl}}\cdot\frac{84.99\text{gNaNO}3}{1\text{molNaNO}3}=18.69\text{gNaNO}3[/tex]

Therefore,

There were formed 31.53g of AgCl (product C) and 18.69g of NaNO3 (product D)


Related Questions

What molarity of a 53.96 mL magnesium hydroxide solution is required to neutralize 93.24mL of a 7.306 M solution of hydrochloric acid solution, which creates as products, magnesium chloride and water?

Answers

Answer

6.312 M

Explanation

Given:

Volume of Mg(OH)₂, Vb = 53.96 mL

Volume of HCl, Va = 93.24 mL

Molarity of HCl, Ma =7.306 M

What to find:

The molarity of Mg(OH)₂, Mb

Solution:

The first step is to write a balanced chemical equation for the reaction.

Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

Mole ratio is 1:2; that is na = 2 and nb = 1

Hence, the molarity of Mg(OH)₂, Mb is calculated using the formula below:

[tex]\frac{M_aV_a}{n_a}=\frac{M_bV_b}{n_b}[/tex]

Plugging the values of the given parameters into the formula, we have:

[tex]\begin{gathered} \frac{7.306M\times93.24mL}{2}=\frac{M_b\times53.96mL}{1} \\ \\ Cross\text{ }multiply \\ \\ M_b\times53.96mL\times2=7.306M\times93.24mL\times1 \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }53.96mL\times2 \\ \\ \frac{M_b\times53.96mL\times2}{53.96mL\times2}=\frac{7.306M\times93.24mL\times1}{53.96mL\times2} \\ \\ M_b=6.312\text{ }M \end{gathered}[/tex]

Therefore, the molarity of a 53.96 mL magnesium hydroxide solution that is required to neutralize 93.24mL of a 7.306 M solution of the hydrochloric acid solution is 6.312 M

Zinc metal reacts with HCl according to the balanced equation:Zn + 2HCl à ZnCl2 + H2When 0.103 g of Zn is combined with enough HCl to make 50 mL of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.5 C to 23.7 C. Find the heat of the reaction. (ans: -160 kJ/mol)

Answers

Answer:

-163kJ/mol

Explanations:

Given the reaction between the zinc metal reacts with HCl according to the balanced equation:

[tex]Zn+2\text{HCl}\rightarrow\text{ZnCl}_2+H_2[/tex]

The required heat of reaction will be calculated using the formula below:

[tex]\triangle H_{\text{rxn}}=-\frac{Q}{n_{Zn}}[/tex]

where:

• Q is the ,heat produced

,

• nZn is the, number of moles that reacted

The formula for calculating the quantity of heat produced is expressed as:

[tex]Q=mc\triangle\theta[/tex]

where:

• m is the ,mass, of the ,Zinc metal

,

• c is the ,specific heat capacity, of zinc

,

• △θ is th,e change in temperature

Get the mass of zinc

[tex]\begin{gathered} \text{mass}=\text{ }density\times\text{volume} \\ m=\rho\times v \end{gathered}[/tex]

The quantity of heat becomes:

[tex]Q=\rho\cdot v\cdot C\triangle\theta[/tex]

Substitute the given parameters to have:

[tex]\begin{gathered} Q=\frac{1.02\cancel{g}}{\cancel{mL}}\times50\cancel{mL}\times\frac{4.18J}{\cancel{g^oC^{}}}\times(23.7-22.5)\cancel{^oC} \\ Q=1.02\times50\times4.18\times1.2 \\ Q=255.8Joules \end{gathered}[/tex]

Next is to get the number of moles of Zinc that reacted (nZn)

[tex]\begin{gathered} nZ_n=\frac{Mass}{\text{Molar mass}} \\ nZ_n=\frac{0.103g}{65.4g\text{/mol}} \\ nZ_n=0.00157mole \end{gathered}[/tex]

Get the required heat of reaction of Zinc:

[tex]\begin{gathered} \triangle H_{\text{rxn}}=-\frac{Q}{nZ_n} \\ \triangle H_{\text{rxn}}=-\frac{255.8Joules}{0.00157moles} \\ \triangle H_{\text{rxn}}=-162,929.93J\text{/mol} \\ \triangle H_{\text{rxn}}\approx-163kJ\text{/mol} \end{gathered}[/tex]

Hence the heat of the reaction of Zinc is approximately -163kJ/mol

0.8g and 1.5g of two metals oxides were found countain 0.64g and 1.2g of metal respectively. show that this results obey law of definite proportion

Answers

Answer

This obey the law of definite proportion

Step-by-step explanation:

The Law of definite proportion states that any chemical compound will always contain a fixed ratio of elements by mass

According to the question,

Let the first oxide = x

Let the second oxide = y

x = 0.8g

y = 1.5g

For the first metal oxide, the amount of metal is 0.64 grams

Hence, the mass of oxygen is (0.8 - 0.64)

Mass of oxygen = 0.16g

Mass ratio = Mass of metal/mass of oxygen

Mass ratio = 0.64 / 0.16

Mass ratio = 4 : 1

For the second metal oxide, the amount of metal is 1.2g

Amount of oxygen = (1.5 - 1.2)

Mass of oxygen = 0.3

Mass of oxygen = 0.3g

Mass ratio = mass of metal/mass of oxygen

Mass ratio = 1.2 / 0.3

Mass ratio = 4: 1

Hence, the ratio of metals to oxygen is in the ratio of 4 to 1

Answer

Hence, this obey the law of definite proportion

What is the wavelength of light with a frequency of 3.215 x 10^15 s-¹?
A. 9.325 x 10^-8m
B. 3.110 x 10^-16m
C. 1.072 x 10^7 m
D. 9.639 x 10^23m

Answers

The wavelength of the light will be 9.63 × 10²³ m .The wavelength of the wave is calculated using the frequency formula.

The quantity of cycles finished in a certain amount of time is known as frequency. It also indicates the number of crests that pass past a specific position in a unit of time. It is often referred to as the reciprocal of time. Hertz are used to express frequency (Hz). The frequency of the wave is calculated using the frequency formula.

The total number of times a repeated event occurs in a unit of time is known as its frequency. Depending on the quantities known, various frequency formulas exist to calculate frequency. The terms frequency (f), time period (T), wave speed (V), and wavelength () are all found using the formula for the frequency of a wave.

We know that ,

Frequency= Speed of light/ wavelength

Wavelength= Frequency × Speed of light

=  3.215 ×10¹⁵ × 2.997× 10⁸

= 9.639 × 10²³ m

The wavelength will be obtained by frequency -wavelength formula a and it will come as 9.63 × 10²³ m .

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A buffer solution is prepared by adding 275 mLof .676 M of HCI to 500 mL of .525M sodiumacetate. What is the pH of this buffer?

Answers

Explanation:

A buffer is a mixture of a weak acid and its conjugate base, or a weak base and its conjugate acid. In our case we are reacting a strong acid (HCl) and a weak base (sodium acetate). So we don't have a buffer. The strong acid will neutralize the weak base. We have to determine which of them is in excess and find the pH of the resulting solution.

HCl (aq) + CH₃COONa (aq) ----> NaCl (aq) + CH₃COOH (aq)

First we have to determine the number of moles of each reactant. We added 275 ml of a 0.676 M solution of HCl and 500 mL of a 0.525 M solution of CH₃COONa. We can use the definition of molarity concentration to determine the number of moles of each reagent.

Molarity = moles of solute/volume of solution in L

moles of solute = molarity * volume of solution in L

moles of HCl = 0.676 M * 0.275 L

moles of HCl = 0.186 moles

moles of CH₃COONa = 0.525 M * 0.500 L

moles of CH₃COONa = 0.262 moles

HCl (aq) + CH₃COONa (aq) ----> NaCl (aq) + CH₃COOH (aq)

In the equation of the reaction all the coefficients are 1. So 1 mol of HCl will completely neutralize 1 mol of CH₃COONa. The molar ratio between them is 1 to 1.

1 mol of HCl = 1 mol of CH₃COONa

We mixed 0.262 moles of CH₃COONa with 0.186 moles of HCl. The 0.186 moles of HCl will neutralize 0.186 moles of CH₃COONa. And CH₃COONa will be excess.

Excess of CH₃COONa = 0.262 moles - 0.186 moles

Excess of CH₃COONa = 0.076 moles

Now we have to determine the concentration of this excess. We mixed 0.275 L with 0.500 L. Then the total volume of solution is 0.775 L. And the concentration of CH₃COONa after the reaction is:

total volume = 0.500 L + 0.275 L

total volume = 0.775 L

Resulting molarity of CH₃COONa = 0.076 moles/0.775 L

Resulting molarity of CH₃COONa = 0.098 M

Finally to get our answer we have to determine the pH of this resulting solution. To determine the pH of a weak base we have to use the ICE table. In solution

Answer:

Okay thanks so much I am watching it and watching

Answers

The twos are the subscript of each of element, those subscripts represent the number of atoms of each element that the molecule has. It means that the correct answer is Subscripts.

You are creating a precipitate of silver chloride, starting with 1.00 mol of silver nitrate (AgNO3) and 0.75 mol of potassium chloride (KCl):AgNO3 (aq) + KCl (aq) → AgCl (s) + KNO3 (aq)What is the maximum amount of silver chloride (AgCl) that can be produced?Select one:a.0.75 molb.0.50 molc.1.00 mold.2.00 mol

Answers

Answer:

a. 0.75 mol

Explanation:

From the balanced equation, we know that 1 mole of AgNO3 reacts with 1 mole of KCl to produce 1 mole of AgCl and 1 mole of KNO3.

Since the ratio between AgNO3 and KCl is 1:1, in this case potassium chloride will be the limiting reactant and silver nitrate will be the excess reactant.

So we have to use the initial moles of the limiting reactant to calculate the moles of product that will be produced.

From the stoichiometry of the reaction we know that from 1 mole of KCl, we can obtain 1 mole of AgCl, so here the ratio between KCl and AgCl is also 1:1, so the maximum amount of silver chloride that can be produced is 0.75moles, as we can see using a mathematical rule of three:

[tex]\begin{gathered} 1molKCl-1moleAgCl \\ 0.75molesKCl-x=\frac{0.75molesKCl*1moleAgCl}{1molKCl} \\ x=0.75molesAgCl \end{gathered}[/tex]

Determine the mole fraction of NaCl in a solution that has 0.589 mol NaCl and 0.625 mol water.

Answers

Step 1

A mole fraction is defined as:

Mole fraction (compound X) = moles of compound X/total moles

It has no units

-------------------------

Step 2

The solution is formed by solute and solvent. Therefore, the total moles will be:

0.589 moles NaCl + 0.625 moles water = 1.214 moles

------------------------

Step 3

Molar fraction (NaCl) = moles NaCl/total moles = 0.589 moles/1.214 moles = 0.49 approx.

Answer: Molar fraction of NaCl = 0.49

Which of the following ranks the molecules OF2, BF2, CF4, in order of their bond angles, from smallest to largest?

Answers

Answer:

[tex]\begin{gathered} BF_3:120\degree \\ CF_4:109.5\degree \\ OF_2:103\degree \\ OF_2

Which statement is true about energy and bonds?A.Energy is absorbed when a bond forms.B.When bonds are formed, energy is released.C.A bond is formed as atoms are split apart from each other.D.Breaking bonds creates energy.

Answers

Answer

B. When bonds are formed, energy is released.

Explanation

The breaking of chemical bonds never releases energy to the external environment. However, energy is only released when chemical bonds are formed.

Therefore the only true statement about energy and bonds in the given options is:

B. When bonds are formed, energy is released.

What is the coefficient of H2O in the balanced half reaction? (See picture)

Answers

Firstly we would determine the oxidation number of the elements to determine which is undergoing oxidation and which is undergoing reduction:

[tex]\begin{gathered} Mn\text{ }in\text{ }MnO_4^-:Mn+(-2\times4)=-1 \\ Mn\text{ }in\text{ }MnO_4^-:Mn-8=-1 \\ Mn\text{ }in\text{ }MnO_4^-:Mn=-1+8=+7 \\ \\ Mn\text{ }in\text{ }MnO_4^{2-}:Mn+(-2\times4)=-2 \\ Mn\text{ }in\text{ }MnO_4^{2-}:Mn=-2+8=+6 \end{gathered}[/tex]

Mn oxidation number goes from +7 to +6 which means it is undergoing a reduction. A decrease in oxidation number means reduction.

[tex]\begin{gathered} S\text{ }in\text{ }HSO_3^-:1+(-2\times3)+S=-1 \\ S\text{ }in\text{ }HSO_3^-:1-6+S=-1 \\ S\text{ }in\text{ }HSO_3^-:S=-1+5=+4 \\ \\ S\text{ }in\text{ }SO_4^{2-}:S+(-2\times4)=-2 \\ S\text{ }in\text{ }SO_4^{2-}:S-8=-2 \\ S\text{ }in\text{ }SO_4^{2-}:S=+6 \end{gathered}[/tex]

S oxidation goes from +4 to +6 meaning it is undergoing oxidation. There is an increase in oxidation number.

We will now balance the oxidation half reaction:

[tex]HSO_3^-+H_2O\rightarrow SO_4^{2-}+3H^++2e[/tex]

The coefficient for the water molecule is 1 in the balnced half reaction.

I have 2.50 x 10²³ atoms of titanium. How many moles of titanium do I have?

Answers

Answer: 2.50 x 10^23 atoms of titanium correspond to 0.415 moles of titanium.

Explanation:

The question requires us to calculate the number of moles of titanium (Ti) that correponds to 2.50 x 10^23 atoms of Ti.

We can apply the Avogadro's number to solve this problem: according to this proportionality constant, there are 6.022 x 10^23 particles in 1 mol of any compound (particles can be atoms, ions, molecules etc).

Thus, considering the Avogadro's number, we can write:

6.022 x 10^23 atoms Ti -------------------- 1 mol Ti

2.50 x 10^23 atoms Ti ---------------------- x

Solving for x, we'll have:

[tex]x=\frac{(1\text{ mol Ti\rparen}\times(2.50\times10^{23}\text{ atoms Ti\rparen}}{(6.022\times10^{23}\text{ atoms Ti\rparen}}=0.415\text{ mol Ti}[/tex]

Therefore, 2.50 x 10^23 atoms of titanium correspond to 0.415 moles of titanium.

can you help me with this, please?which anion will form a precipitate with Na+?

Answers

A precipitate formed in a reaction or solution is when we have a solid salt that is not soluble in the reaction anymore, in our case we are dealing only with the compound formed with Sodium and another anion, so the one that is not soluble in water will be the one that will form a precipitate, therefore a solid salt not dissolvable. Out of the three ionic compounds: Na2CO3, Na2CrO4, NaI2, Sodium chromate (Na2CrO4) will present as the less soluble in water, since chromates in general are not very soluble in water, which is not the case for carbonates and iodides, they will readily dissolve in water, not forming any precipitates

Using the following chemical equation and your vast knowledge of stoichiometry, calculate the exact molarity of the NaOH. Hint—it should be around 0.1 M.KHP(aq) + NaOH(aq) NaKP(aq) + H2O(liquid)a)Convert the grams of KHP into moles of KHP using the formula weight of 204.2 grams per mole.b)Convert moles of KHP into moles of NaOH using the stoichiometry given in the above equation.c)Calculate the molar concentration of the NaOH by dividing the number of moles of NaOH calculated immediately above by the volume of NaOH used (measured using the buret). Be sure to convert the volume from mL to L.d)Average the three values obtained to determine the average molar the concentration of the NaOH. GET INFORMATION FROM THE PICTURE

Answers

ANSWER

The molar concentration of NaOH is 0.0705 mol/L

STEP-BY-STEP EXPLANATION:

Given the balanced equation below

[tex]\text{KHP}_{(aq)}+NaOH_{(aq)}\text{ }\rightarrow NaKP_{(aq)}+H_2O_{(l)}[/tex]

According to the balanced equation, 1 mole of KHP gives 1 mole of NaOH

Given parameters

Molar mass of KHP = 204.2 grams/mol

To find the mole of KHP, we will need to find the average grams of KHP used

• For flask 1; 0.55g of KHP was used

,

• For flask 2; 0.56g of KHP was used

,

• For flask 3; 0.56g of KHP was used

The average mass of KHP used can be calculated below using the average formula

[tex]\begin{gathered} \text{Average mass = }\frac{mass\text{ 1 + mass 2 + mass 3}}{3} \\ \text{Average mass = }\frac{0.55\text{ + 0.56 + 0.56}}{3} \\ \text{Average mass = }\frac{1.166}{3} \\ \text{Average mass = 0.3886 grams} \end{gathered}[/tex]

The average mass of KHP used is 0.3886grams

[tex]\begin{gathered} \text{Mole = }\frac{\text{ reacting mass}}{\text{molar mass}} \\ \text{reacting mass = 0}.3886\text{ grams} \\ \text{Molar mass = 2}04.2\text{ grams/mol} \\ \text{Mole = }\frac{0.3866\text{ }}{204.2} \\ \text{Mole of KHP = 0.0019 mole} \end{gathered}[/tex]

The mole of KHP is 0.0019 mole

PART B

According to the balanced equation, the stoichiometry ratio of KHP to NaOH is 1: 1

Let the mole of NaOH be x

[tex]\begin{gathered} 1\text{ : 1 = }0.0019\text{ : x} \\ \frac{1}{1}\text{ = }\frac{0.0019}{x} \\ \text{Cross multiply} \\ 1\cdot\text{ x = 1 }\cdot\text{ 0.0019} \\ x\text{ = 0.0019 mole} \end{gathered}[/tex]

Hence, the mole of NaOH is 0.0019 mole

PART C

Given the following parameters

0. The volume of NaOH used in flask 1 = 26.70mL

,

1. The volume of NaOH used in flask 2= 27.09mL

,

2. The volume of NaOH used in flask 3 = 26.96mL

The next step is to convert the mL to L

[tex]1mL\text{ = 0.001L}[/tex]

For flask 1

[tex]\begin{gathered} 1mL\text{ = 0.001L} \\ \text{Let x be the volume of NaOH in L} \\ \text{ 1mL = 0.001L} \\ 26.70mL\text{ = xL} \\ \text{Cross multiply} \\ xL\cdot\text{ 1ml = 26.70mL }\cdot\text{ 0.001L} \\ x\text{ = }\frac{26.70\cdot\text{ 0.001}}{1} \\ x\text{ = 0.0267L} \end{gathered}[/tex]

Using the same conversion process

The volume of NaOH in L in flask 2 = 0.02709L

The volume of NaOH in L in flask 3 = 0.02696L

Hence, the molar concentration of the solution in each flask can be calculated as follows

[tex]\text{Molar concentration = }\frac{concentration}{\text{Volume}}[/tex]

For flask 1

Mole of NaOH = 0.0019 mole

Volume of NaOH = 0.0267L

[tex]\begin{gathered} \text{Molar concentration = }\frac{0.0019}{0.0267} \\ \text{Molar concentration = }0.0711\text{ mol/L} \end{gathered}[/tex]

For flask 2

Mole = 0.0019 mole

Volume = 0.02709L

[tex]\begin{gathered} \text{Molar concentration = }\frac{Concentration}{\text{volume}} \\ \text{Molar concentration = }\frac{0.0019}{0.02709} \\ \text{Molar concentration = 0.0701 mol/L} \end{gathered}[/tex]

For flask 3

Mole = 0.019mole

Volume = 0.02696 L

[tex]\begin{gathered} \text{Molar concentration = }\frac{concentration}{\text{volume}} \\ \text{Molar concentration = }\frac{0.0019}{0.02696} \\ \text{Molar concentration = 0.0704 mol/L} \end{gathered}[/tex]

PART D

Average molar concentration can be found using the below formula

[tex]\begin{gathered} \text{Average molar concentration = }\frac{0.0711\text{ + 0.0701 + 0.0704}}{3} \\ \text{Average molar concentration =}\frac{0.2116}{3} \\ \text{Average molar concentraion = 0.0705 mol/L} \\ \text{The average molar concentration of NaOH is 0.0705 mol/L} \end{gathered}[/tex]

Is chlorine malleable or brittle?

Answers

Chlorine is not malleable or ductile. This is so because nonmetals, like chlorine, are brittle.

What is a chlorine physical characteristic?

The odor of chlorine is distinctively intrusive and upsetting. The liquid is pure amber, while the gas is a greenish yellow tint. The information on the physical characteristics of chlorine as discovered by various researchers reveals some variances.

What is brittle versus ductile or malleable?

In essence, metals that are malleable or ductile can both deform in some ways through plastic deformation in response to stress. Brittle metals are those that fracture without significant plastic deformation. Brittle is the antithesis of ductile and malleable in this context.

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How many grams of He gas are in a 33.2L container at 1.13atm and 652K?

Answers

Step 1

The He gas is assumed to be ideal. Therefore, the next equation is applied:

[tex]p\text{ x V = n x R x T \lparen1\rparen}[/tex]

----------------------

Step 2

Information provided:

p = pressure = 1.13 atm

V = volume = 33.2 L

n = number of moles = unknown

T = absolute temperature = 652 K

R = gas constant = 0.082 atm L/mol K

---------------------

Step 3

Firstly, n is calculated from (1):

[tex]\begin{gathered} \frac{p\text{ x V}}{R\text{ x T}}=\text{ n} \\ \frac{1.13\text{ atm x 33.2 L}}{0.082\text{ }\frac{atm\text{ L}}{mol\text{ K}}x652\text{ K}}=0.702\text{ moles} \end{gathered}[/tex]

n = 0.702 moles

----------------------

Step 4

The mass of He is calculated as:

n = mass/the molar mass He

The molar mass of He = 4.00 g

So, n = mass/4.00 g/mol

n x 4.00 g/mol = mass

0.702 moles x 4.00 g/mol = 2.81 g

Answer: mass = 2.81 g

17. What is a bond?

Answers

Bond:

Bonds are forces that keep atoms together to make compounds or molecules.

In chemical bonds electrons are involved. And there are three kinds of bonds:

1. Covalent bonds: where the electrons are shared between the atoms.

2. Polar covalent bonds

3. Ionic bonds: happens when one atom gives to the other atom electrons.

14) How many mL of 5.724 M HNO3 are needed to prepare a 183.95 mL of 2.063 M HNO3?

Answers

Answer

66.30 mL

Explanation

Given parameters:

Initial concentration, C₁ = 5.724 M

Final volume, V₂ = 183.95 mL

Final volume, V₂ = 183.95 mLFinal concentration, C₂ = 2.063 M

What to find:

The initial volume, V₁

Step-by-step solution:

Using the dilution law:

[tex]C_1V_1=C_2V_2[/tex]

Substitute the given parameters into the formula to get V₁

[tex]\begin{gathered} 5.724M\times V_1=2.063M\times183.95mL \\ \text{Divide both sides by 5.724M} \\ \frac{5.724M\times V_1}{5.724M}=\frac{2.063M\times183.95mL}{5.724M} \\ V_1=66.30\text{ mL} \end{gathered}[/tex]

Hence, 66.30 mL of 5.724 M HNO3 are needed to prepare a 183.95 mL of 2.063 M HNO3?

What affect does a chemical reaction have on atoms?

Answers

Chemical reactions involve breaking chemical bonds between reactant molecules (particles) and forming new bonds between atoms in product particles (molecules). The number of atoms before and after the chemical change is the same but the number of molecules will change.

Answer:Chemical reactions involve breaking chemical bonds between reactant molecules (particles) and forming new bonds between atoms in product particles (molecules).Explanation:  

[OH﹘] = 6.5 x 10-5Mfind the pH and pOH

Answers

Answer:

pH = 9.81

pOH = 4.19

Explanations:

The formula for calculating the pOH of a solution is given as:

[tex]pOH=-log[OH^-][/tex]

Given the following parameters

[tex][OH^-]=6.5\times10^{-5}M[/tex]

Substitute

[tex]\begin{gathered} pOH=-log(6.5\times10^{-5}) \\ pOH=-(-4.19) \\ pOH=4.19 \end{gathered}[/tex]

Determine the pH of the solution

[tex]\begin{gathered} pH+pOH=14 \\ pH=14-pOH \\ pH=14-4.19 \\ pH=9.81 \end{gathered}[/tex]

Therefore the pH and pOH are 9.81 and 4.19 respectively

4.68g of Si and 5.32g of O what is the mass percentage composition

Answers

According to the what was presented to me, to find the number of moles of a compound or element from a given mass, we need to use the molar mass of this element and do the following calculation:

Si = 28.0855g/mol

28.08g = 1 mol of Si

4.68g = x moles of Si

x = 0.17 moles of Si

O = 16g/mol

16g = 1 mol of O

5.32g = x moles of O

x = 0.33 moles of O

A 45.0 ml sample of nitric acid is neutralized by 119.4 ml of 0.2M NaOH. What is the molarity of the nitric acid?

Answers

Answer:

0.53L

Explanations:

According to dilution formula:

[tex]C_1V_1=C_2V_2[/tex]

C1 and C2 are the initial and final concentrations

V1 and V2 is initial and final volume

Given the following parameters

• V1 = 45mL

,

• C2 = 0.2M

,

• V2 = 119.4mL

Substitute the given parameters into the formula

[tex]\begin{gathered} C_1=\frac{C_2V_2}{V_1} \\ C_1=\frac{0.2\times119.4}{45} \\ C_1=\frac{23.88}{45} \\ C_1=0.53M \end{gathered}[/tex]

Hence the molarity of the nitric acid is 0.53L

Determine the mass of the sample based on the following data to two decimal places- mass of beaker and cover: 50.09- mass of beaker, cover, and sample before heating: 51.04- mass of beaker, cover, and sample after heating: 50.88

Answers

Step 1 - Understanding the problem

We need to determine the mass of a sample which probably contains water as an impurity. Therefore, if we just weight the sample, we won't be able to separate its own weight from that of water.

What can be done, then, is heating the sample first in order to evaporate all the water, and only then weighting it. That's exactly what the procedure in the exercise has done.

The idea here thus is that we just need to subtract two weights:

[tex]m_{\text{sample}}=m_{\text{sample}+\text{water}}-m_{\text{water}}[/tex]

Step 2 - Discovering the mass of the sample

We can discover the mass of the sample by simply subtracting the mass of beaker and cover from the final mass (beaker + cover + sample after heating).

This will give us the mass of the sample because the water has evaporated after heating:

[tex]m_{\text{sample}}=50.88-50.09=0.79g[/tex]

The mass of the sample is thus 0.79g.

How must nuclear waste that is considered high-level waste to be stored

Answers

Answer: the best option to answer the question is the second one (letter B).

Explanation:

The question requires us to choose, among the options given, which one best corresponds to the correct form of disposal fo high-level nuclear waste.

High-level nuclear waste corresponds mainly to highly radioactive materials produced as byproducts of reactions that occur inside nuclear reactors. When they are not of use anymore, they are disposed in underground repositories, without any recycling. The place of disposal of this type of nuclear waste is usually placed in cannister which are placed underground and sealed with rocks. Since the time of decay of high-level nuclear waste can reach hundreds of thousands of years, this type of waste must be disposed of in a way that provides adequate protection for a long time.

Considering the information above, we can say that the best option to answer the question is the second one (letter B).

you discover the head of a match contains 3.75 g of sulfur. How many atoms of sulfur does the match contain? (hint: grams > moles > atoms)

Answers

As the question says, we are going to follow the path of the hint, first by finding the number of moles of Sulfur in 3.75 grams, and we can do that by using its molar mass, which is 32g/mol

32g = 1 mol

3.75g = x moles

32x = 3.75

x = 3.75/32

x = 0.12 moles of Sulfur in 3.75 grams

Now we have the number of moles, and to find the number of atoms, we need to use the Avogadro's constant number, which is the number of atoms in a single mol, this value is 6.02*10^23 atoms in 1 mol

1 mol = 6.02*10^23 atoms

0.12 moles = x atoms

x = 7.22*10^22 atoms of Sulfur in 3.75 grams

How many atoms (or molecules) are present in 1 mole?Question 19 options:A) 1B) 6.022×1020C) 6.022×1023D) 6.022

Answers

Answer:

The answer is C.

Explanation:

The number of particles (atoms/molecules) in 1 mole is referred to as the Avogadro's number. The Avogadro's number = 6.022 * 10 ^ 23. Therefore, the answer is C.

The frequency of a wave is 1.2 x 10° Hz. What is its wavelength? What kind of wave is this?

Answers

Answer: wavelength=  ............

Explanation:

Are all covalent molecules soluble in water ?

Answers

Answer: Covalent molecules are one way in which they soluble in anything
Explanation:
determining which kind of elements are soluble, especially covalent molecules. Chemical balances, no matter in bonds. chemical balances, physical balances are reasons why.

Calculate the percentage oxygen in dinitrogen pentoxide

Answers

So, To calculate the mass percentage of dinitrogen pentoxide (N2O5) we will first calculate the molar weight of the molecule. The mass percentage will be calculated with the following equation:

[tex]\text{Mass percentage = }\frac{\text{Molar Mass oxygen}}{MolarmassmoleculeN_2\text{O}_5}\times100[/tex]

Molar mass will be:

Element Atomic Mass #Atoms Mass

N 14.0067 2 28.0134 g/mol

O 15.9994 5 79.997 g/mol

Total mass = 28.0134 + 79.997 = 108.0104 g/mol

[tex]\text{Mass percentage = }\frac{\text{7}9.997}{108.0104}\times100\text{ = }74.064\text{ percent}[/tex]

9. I have a pressure of 65 kPa and a starting temperature of29°C. If I raise the temperature to 167°C, what is the newpressure?

Answers

Answer:

[tex]94.70\text{ kPa}[/tex]

Explanation:

Here, we want to get the new pressure

According to the pressure law, temperature and pressure are directly proportional

Mathematically, we have that as:

[tex]\text{ }\frac{P_1}{T_1}\text{ = }\frac{P_2}{T_2}[/tex]

where:

P1 is the initial pressure which is 65 KPa

P2 is the final pressure that we want to calculate

T1 is the initial temperature which we will convert to Kelvin by adding 273K:

We have T1 as 29 + 273 = 302K

T2 is the final temperature which is also 167 + 273 = 440K

Substituting the values, we have it that:

[tex]\begin{gathered} \frac{65}{302}\text{ = }\frac{P_2}{440} \\ \\ P_2\text{ = }\frac{440\times65}{302}\text{ = 94.70 kPa} \end{gathered}[/tex]

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