Answer:
158 L.
Explanation:
What is given?
Pressure (P) = 1 atm.
Temperature (T) = 112 °C + 273 = 385 K.
Mass of methane CH4 (g) = 80.0 g.
Molar mass of methane CH4 = 16 g/mol.
R constant = 0.0821 L*atm/mol*K.
What do we need? Volume (V).
Step-by-step solution:
To solve this problem, we have to use ideal gas law: the ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. The formula is:
[tex]PV=nRT.[/tex]Where P is pressure, V is volume, n is the number of moles, R is the constant and T is temperature.
So, let's find the number of moles that are in 80.0 g of methane using its molar mass. This conversion is:
[tex]80.0g\text{ CH}_4\cdot\frac{1\text{ mol CH}_4}{16\text{ g CH}_4}=5\text{ moles CH}_4.[/tex]So, in this case, n=5.
Now, let's solve for 'V' and replace the given values in the ideal gas law equation:
[tex]V=\frac{nRT}{P}=\frac{5\text{ moles }\cdot0.0821\frac{L\cdot atm}{mol\cdot K}\cdot385K}{1\text{ atm}}=158.04\text{ L}\approx158\text{ L.}[/tex]The volume would be 158 L.
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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?
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
Which of the following ranks the molecules OF2, BF2, CF4, in order of their bond angles, from smallest to largest?
Answer:
[tex]\begin{gathered} BF_3:120\degree \\ CF_4:109.5\degree \\ OF_2:103\degree \\ OF_2Using 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
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]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?
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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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
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]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?
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]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?
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
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?
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
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
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
Examine the table According to the table, which reaction is correct.
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).
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?
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 mole fraction of NaCl in a solution that has 0.589 mol NaCl and 0.625 mol water.
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
What is the coefficient of H2O in the balanced half reaction? (See picture)
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.
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
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.
Are all covalent molecules soluble in water ?
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?
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.
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.
Percent is the amount of one part of a mixture per 1000 total parts.A)trueB)false
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.
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
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
when is standard reduction potential decreasing
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.
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
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
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What affect does a chemical reaction have on atoms?
Answer:Chemical reactions involve breaking chemical bonds between reactant molecules (particles) and forming new bonds between atoms in product particles (molecules).Explanation:
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
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.
How must nuclear waste that is considered high-level waste to be stored
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).
4.68g of Si and 5.32g of O what is the mass percentage composition
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
How many atoms (or molecules) are present in 1 mole?Question 19 options:A) 1B) 6.022×1020C) 6.022×1023D) 6.022
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.
I have 2.50 x 10²³ atoms of titanium. How many moles of titanium do I have?
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+?
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
17. What is a bond?
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.