Answers
Answer:
Hafnium is more likely to be the identity of the substance.
Explanation:
The given information about the substance is:
- Mass: 46.9g
- Volume: 3.5cm^3
- Solid at room temperature (23°C)
We can calculate the density of the substance by replacing the values of mass and volume in the density formula:
[tex]\begin{gathered} Density=\frac{mass}{volume} \\ . \\ Density=\frac{46.9g}{3.5cm^3} \\ . \\ Density=13.4\frac{g}{cm^3} \end{gathered}[/tex]Now we know that the density of the substance is 13.4g/cm^3.
At this point, the substance could be mercury or hafnium, because they have a similar density, but we also know that the substance is solid at 23°C, so since hafnium has a melting point of 2,233°C, it is more likely to be the identity of the substance.,
Related Questions
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]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
veredAnswerQuestion 3Suppose the gas inside a closed, steel cylinder has a pressure of 6.1 atm at2.0 °C. What will be the pressure (atm) of the gas at -83.0 °C? Enter youranswer to 2 decimal places.-710/1 pts4.22 margin of error +/- 0.05
Answers
Answer
4.22 atm
Explanation
Given that:
The initial pressure, P₁ = 6.1 atm
The initial temperature, T₁ = 2.0 °C = (2.0 + 273.15 K) = 275.15 K
The final temperature, T₂ = -83.0 °C = (-83.0 + 273.15 K) = 190.15 K
What to find:
The final pressure, P₂ (atm) of the gas at 83.0 °C.
Step-by-step solution:
The relationship between pressure and temperature of gaseous subtance was decribed by Amonton's gas law.
Amonton's Law states that the pressure of an ideal gas varies directly with the absolute temperature when the volume of the sample is held constant.
Mathematically, we have:
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]Putting the values of the parameters into the formula, we have:
[tex]\begin{gathered} \frac{6.1\text{ }atm}{275.15\text{ }K}=\frac{P_2}{190.15\text{ }K} \\ \\ P_2=\frac{6.1atm\times190.15K}{275.15K} \\ \\ P_2=4.22\text{ }atm \end{gathered}[/tex]Thus, the final pressure, (atm) of the gas at 83.0 °C is 4.22 atm.
Examine the table According to the table, which reaction is correct.
Answers
Answer:
Third option, because copper (Cu) is a stronger oxidizing agent than zinc (Zn).
Explanation:
The table shows the Oxidation Half-Reaction, so the elements oxidizes.
When an element is oxidized, it causes the other element in the reaction to be reduced, therefore the element being oxidized is a reducing agent.
In this case, the correct reaction is:
[tex]Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu[/tex]because, copper (Cu) is a stronger oxidizing agent than zinc (Zn).
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
Consider the following redox reaction: 2Ag+(aq) + Zn(s) --> Zn^2+ (s) + 2Ag(s)Which of the following statements are TRUE? (Select All That Apply)a) Ag+ is the oxidizing agentb) Ag+ is reduced to Agc) Zn oxidize to Zn^2+d) Zn is the reducing agente) Zn reduce to Zn^2+d) Ag+ oxidize to Zn^2+
Answers
ANSWER
Ag+ is the oxidizing agent
Ag+ is reduced to Ag
Zn is oxidized to zn^2+
Zn is the reducing agent
option ABCD
STEP-BY-STEP EXPLANATION:
Given information
[tex]2Ag^+_{(aq)}+Zn_{(s)\text{ }\rightarrow\text{ }}Zn^{2+}_{(s)}+2Ag_{(s)}[/tex]From the above redox reaction, you will see that there is a change in the oxidation number of zinc and silver.
Zn goes from 0 to 2+ in its oxidation state
Hence, zinc has been oxidized, therefore, it is a reducing agent
Ag goes from +1 to 0 in its oxidation state
Hence, Ag has been reduced, therefore, it is an oxidizing agent
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.
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.
2. The same ratios can be used to predict the mass of the products given the mass of the reactants.a. A sample of copper (II) sulfate pentahydrate weighs 5.9 grams.How many grams of water would be released if all of the water is removed from the pentahydrate?Pls see pictures for details
Answers
Answer
2.12872 grams of water
Explanation
Given:
Mass of copper (II) sulfate pentahydrate sample = 5.9 grams
What to find:
The grams of water that would be released if all of the water is removed from the 5.9 grams sample of copper (II) sulfate pentahydrate.
Step-by-step solution:
The chemical formula of copper (II) sulfate pentahydrate is: CuSO₄.5H₂O
Molar mass of CuSO₄.5H₂O = 249.68 g/mol
Mass of water in 1 mole of CuSO₄.5H₂O = 5 x molar mass of water = 5 x 18.01528 g/mol = 90.0764 g/mol
the percentage by mass of water in 1 mole of CuSO₄.5H₂O is
[tex]\begin{gathered} \%\text{ }by\text{ }mass\text{ }of\text{ }water=\frac{Mass\text{ }of\text{ }water}{Molar\text{ }mass\text{ }of\text{ }CuSO₄.5H₂O}\times100\% \\ \\ \%\text{ }by\text{ }mass\text{ }of\text{ }water=\frac{90.0764}{249.68}\times100\% \\ \\ \%\text{ }by\text{ }mass\text{ }of\text{ }water=36.08\% \end{gathered}[/tex]Hence, you can now use the % by mass of water in 1 mole of copper (II) sulfate pentahydrate to determine the mass of water that would be released in 5.9 grams of copper (II) sulfate pentahydrate as shown below.
[tex]\begin{gathered} Mass\text{ }of\text{ }water=\frac{36.08}{100}\times5.9\text{ }grams \\ \\ Mass\text{ }of\text{ }water=0.3608\times5.9grams \\ \\ Mass\text{ }of\text{ }water=2.12872\text{ }grams \end{gathered}[/tex]Therefore, the grams of water that would be released if all of the water is removed from the 5.9 grams of copper (II) sulfate pentahydrate is 2.12872 grams
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.
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
What is the percent by mass concentration of salt in a solution made by dissolving 3.5 g of salt in 75 kg of water?Please include equation used and variable you solved for.
Answers
Answer:
Explanations:
The formula for calculating the percentage by mass of concentration is expressed as;
[tex]undefined[/tex]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 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.
Suppose 3.04 g of copper is reacted with excess nitric acid according to the given equation. Cu(s) + 4NHO3(aq) ---> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(1)
What is the theoretical yield of Cu (NO3)2?
Answers
Taking into account the reaction stoichiometry, the theoretical yield of Cu(NO₃)₂ is 8.9717 grams.
Reaction stoichiometryIn first place, the balanced reaction is:
Cu + 4 NHO₃ → Cu(NO₃)₂ + 2 NO₂ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
Cu: 1 moleNHO₃. 4 molesCu(NO₃)₂: 1 moleNO₂: 2 molesH₂O: 2 molesThe molar mass of the compounds is:
Cu: 63.55 g/moleNHO₃: 63 g/moleCu(NO₃)₂: 187.55 g/moleNO₂: 62 g/moleH₂O: 18 g/moleThen, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
Cu: 1 mole ×63.55 g/mole= 63.55 gramsNHO₃: 4 moles ×63 g/mole= 252 gramsCu(NO₃)₂: 1 mole ×187.55 g/mole= 187.55 gramsNO₂: 2 moles ×62 g/mole= 124 gramsH₂O: 2 moles ×18 g/mole= 36 gramsTheoretical yieldThe theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
Theoretical yield of Cu(NO₃)₂The following rule of three can be applied: if by reaction stoichiometry 63.55 grams of Cu form 187.55 grams of Cu(NO₃)₂, 3.04 grams of Cu form how much mass of Cu(NO₃)₂?
mass of Cu(NO₃)₂= (3.04 grams of Cu× 187.55 grams of Cu(NO₃)₂)÷ 63.55 grams of Cu
mass of Cu(NO₃)₂= 8.9717 grams
Finally, the maximum amount of Cu(NO₃)₂ is 8.9717 grams.
Learn more about the reaction stoichiometry:
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Percent is the amount of one part of a mixture per 1000 total parts.A)trueB)false
Answers
The word percent is telling us 'percent' is the amount of one part in every hundred (100), so the given statement of the question is false.
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
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
12. Which two elements have chemical properties that are most
similar?
A.C and N
B. Cl and Ar
C.K and Ca
D.Li and Na
Answers
Li and Na two elements have chemical properties that are most similar
The element that have the most similar chemical properties are those in the same group or column of the periodic table and one such group includes lithium sodium potassium these element are all shiny and conduct heat and electricity well and have similar chemical properties and the element in a group have the same number of electron in the outermost shell hence they have similar chemical properties
Know more about element
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Which element is in the same group as Lithium (Li)?
Carbon (C)
Potassium (K)
Chlorine (Cl)
Hydrogen (H)
Answers
Answer: Potassium (K)
Explanation:
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_2Determine 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
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.
when is standard reduction potential decreasing
Answers
In general, the ions of very late transition metals those towards the right-hand end of the transition metal block, such as copper, silver and gold have high reduction potentials. In other words, their ions are easily reduced.
Explanation:Again, note that when calculating E∘cell E cell ∘ , standard reduction potentials always remain the same even when a half-reaction is multiplied by a factor. Standard electrode potential - When the concentration of all the species is taken as unity in the half-cell reaction, then the calculated electrode potential is called the standard electrode potential. The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive values of reduction potential are indicative of a greater tendency to get reduced. The electrode potential depends upon the concentrations of the substances, the temperature, and the pressure in the case of a gas electrode.
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]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]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:
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.
A lead block, initially at 58.5oC, is submerged into 100.0 g of water at 24.8oC, in an insulated container. The final temperature of the mixture upon reaching equilibrium is 26.2oC. What is the mass of the lead block?
Answers
Answer:
The specific heat capacity of lead is 0.128 J/g°C.
Mass of lead block = (100.0 g)(26.2oC - 24.8oC)(0.128 J/g°C) / (58.5oC - 24.8oC)
Mass of lead block = 5.45 g
Determine the volume of hydrogen gas needed to react completely with 6.00 L of oxygen gas to form water. (20pts)2H2(g) + O2(g) ⇨ 2H2O(g) at STP
Answers
Explanation
Given
2H2(g) + O2(g) ⇨ 2H2O(g)
Volume = 6.00 L
At STP: Temperature = 273K
Pressure = 1 atm
We know that R constant = 0.0821 L.atm/mol.K
Solution
Step 1: Calculate the number of moles of oxygen at STP
Use the ideal gas law equation:
PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature.
n = PV/RT
n = (1 atm x 6.00 L)/(0.0821 L.atm/mol.K x 273 K)
n = 0.27 mol
Step 2: Use the stoichiometry to find the moles of H2
The molar ratio between oxygen and hydrogen is 1:2
Therefore the moles of H2 = 0.27 x (2/1) = 0.535 mol
Step 3: Calculate the volume of H2
V = nRT/P
V = (0.535 mol x 0.0821 L.atm/mol.K x 273 K)/1 atm
V = 12.0 L
Answer
Volume of hydrogen = 12.0 L
Aspirin is a common analgesic. If you want to produce 500. mg of aspirin (C9H8O4, m.w. = 180.16 g/mol) from the reaction of C7H6O3 (m.w. = 130.12 g/mol) and C4H6O3 (m.w. = 102.09 g/mol), what is the minimum amount of C4H6O3 that is needed?2C7H6O3(s) + C4H6O3(l) → 2C9H8O4(s) + H2O(l)
Answers
1) First, let's rewrite the chemical equation here:
2 C7H6O3(s) + C4H6O3(l) → 2 C9H8O4(s) + H2O(l)
2) Let's find out how many moles of C9H8O4 have into 500 mg of it. For this, use ese the following equation:
mole = mass/ molar mass
mass = 500 mg = 0.5 g
molar mass of C9H8O4 = 180.16 g/mol (the question give it to us)
mole = 0.5/180.16
mole = 0.0028 moles of C9H8O4
3) Using the chemical equation proportion, we can discover the amount in moles of C4H6O3 needed to form 0.0028 moles of C9H8O4. So:
1 mol of C4H6O3 forms 2 moles of C9H8O4
x mol of C4H6O3 forms 0.0028 moles of C9H8O4
2x = 0.0028
x = 0.0014 moles of C4H6O3
4) We need to find the answer in grams. Let's transform 0.0014 moles of C4H6O3 into grams. For this, use ese the following equation:
mass = mole × molar mas
mass = 0.0014 moles × 102.09
mass = 0.142 grams or 1.42x10⁻¹
Answer: The amount of C4H6O3 that is needed is 1.42x10⁻¹ g.