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2 years ago

Question 1(Multiple Choice Worth 3 points)

Chemistry

1 answer:

Bingel [31]2 years ago

4 0

1) Answer:

moles of oxygen (n) = 1.86 moles

Explanation:

according to Boyle's law the formula to solve this problem is:

PV=nRt

when P is the pressure which equal 1.25 atm

and V is the volume which equal 37.5 L

n is the number of moles which we need to calculate it

R is constant which equal 0.082

t is the temperature in kelvin

By substitution:

1.25*37.5 = n * 0.082 * 307

So n = 1.86 moles

2) Answer:

the volume of oxygen gas = 34 L

Explanation:

at standard temperature and pressure (STP) 1 mole of gas will equal = 22.4L

So when we have 1.5 moles of oxygen at standard temperature and pressure (STP) so we will estimate it like that

1.5 moles *22.4 L/ 1 mole = approximately 34 L

3) Answer:

The volume of H2 = 2.29 L

Explanation:

according to the Balanced equation we can see that the molar ratio between Zn and H2 1 : 1

so to know the number of moles of H2 we will get it for Zn first :

number of moles Zn = mass of Zn / molar mass Zn

= 5.98 / 65.39 =0.0914 moles

so number of moles H2 = 0.09 moles

by substitution in the following formula:

PV = nRT

0.978 * V = 0.09 * 0.082 * 298

so The volume of H2 = 2.29 L

4) Answer:

Volume of O2 = 1.4 L

Explanation:

first we have to balance the equation:

2Na2O2 +2CO2 → 2Na2CO3 + O2

2 mole CO2 give 1 mole of O2 so the molar ratio is 2:1

at STP 1 mole of gas will equal = 22.4 L

??? moles of CO2 = 2.8

n CO2 = 0.125 moles so n O2 = 0.125 /2 = 0.0625 moles

so when 1 mole of as = 22.4

0.0625 moles O2 = ???

Volume of O2 =0.0625 moles * 22.4 L/ 1 mole

= 1.4 L

5) Answer:

the initial quantity of sodium metal used = 17.2 gram

Explanation:

at STP 1 mole of gas will equal = 22.4 L

so moles of H2 equal ?? when 8.40 liters of H2 gas were produced

so moles of H2 = 8.4/22.4 =0.375 moles

and according to the balanced equation the molar ratio between H2 ans Na is 1 : 2

so number of moles for Na = 0.375 *2 = 0.75 moles

to get the initial quantity of sodium metal (mass Na) = number of moles * molar mass

mass Na = 0.75 moles * 23 gm/mole= 17.25

6) Answer:

False

Explanation:

because STP means standard temperature and pressure. and the Standard temperature must be 273 K and the standard pressure must be 1 atm but in the question the temperature is 298 K not 273 K so , It is not a standard temperature

7) Answer:

22.4 liters

Explanation:

1 mole of gas will equal = 22.4 L

because at the Standard temperature must be 273 K and the standard pressure must be 1 atm

so V = nRT/P

=1 mole * 0.082 * 273 K / 1 aTm

= 22.4 liters

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IrinaVladis [17]

Answer:

34.2 g is the mass of carbon dioxide gas one have in the container.

Explanation:

Moles of $O_2$ :-

Mass = 49.8 g

Molar mass of oxygen gas = 32 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{49.8\ g}{32\ g/mol}$

$Moles_{O_2}= 1.55625\ mol$

Since pressure and volume are constant, we can use the Avogadro's law as:-

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

Given ,

V₂ is twice the volume of V₁

V₂ = 2V₁

n₁ = ?

n₂ = 1.55625 mol

Using above equation as:

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

$\frac {V_1}{n_1}=\frac {2\times V_1}{1.55625}$

n₁ = 0.778125 moles

Moles of carbon dioxide = 0.778125 moles

Molar mass of $CO_2$ = 44.0 g/mol

Mass of $CO_2$ = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g

34.2 g is the mass of carbon dioxide gas one have in the container.

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2 years ago

A sample of oxygen gas was found to effuse at a rate equal to two times that of an unknown gas. what is the molar mass (in g/mol

inn [45]

128 g/mol Using Graham's law of effusion we have the formula: r1/r2 = sqrt(m2/m1) where r1 = rate of effusion for gas 1 r2 = rate of effusion for gas 2 m1 = molar mass of gas 1 m2 = molar mass of gas 2 Since the atomic weight of oxygen is 15.999, the molar mass for O2 = 2 * 15.999 = 31.998 Now let's subsitute the known values into Graham's equation and solve for m2. r1/r2 = sqrt(m2/m1) 2/1 = sqrt(m2/31.998) 4/1 = m2/31.998 127.992 = m2 So the molar mass of the unknown gas is 127.992 g/mol. Rounding to 3 significant figures gives 128 g/mol

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2 years ago

In a particular mass of kau(cn)2, there are 6.66 × 1020 atoms of gold. What is the total number of atoms in this sample?

Vinvika [58]

Total number of atoms in the sample is the sum of number of atoms of all the elements present in the sample.

Number of gold, $Au$ atoms in $KAu(CN)_2$ = $6.66\times 10^{20}$ atoms (given)

From the formula of compound that is $KAu(CN)_2$ it is clear that the number of potassium and gold are same whereas those of carbon and nitrogen are 2 times of them.

So, the number of atoms of each element is:

Number of potassium, $K$ atoms in $KAu(CN)_2$ = $6.66\times 10^{20}$ atoms

Number of carbon, $C$ atoms in $KAu(CN)_2$ = $2\times6.66\times 10^{20}$ atoms = $13.32\times 10^{20}$

Number of nitrogen, $N$ atoms in $KAu(CN)_2$ = $2\times6.66\times 10^{20}$ atoms = $13.32\times 10^{20}$

Total number of atoms in $KAu(CN)_2$ = Number of gold, $Au$ atoms+Number of potassium, $K$ atoms +Number of carbon, $C$ atoms + Number of nitrogen, $N$ atoms

Total number of atoms in $KAu(CN)_2$ = $6.66\times 10^{20}+6.66\times 10^{20}+13.32\times 10^{20}+13.32\times 10^{20}$

Total number of atoms in $KAu(CN)_2$ = $39.96\times 10^{20}$ atoms

Hence, the total number of atoms in $KAu(CN)_2$ is $3.996\times 10^{21}$ atoms.

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2 years ago

A 8.6 g sample of methane and 15.6 g sample of oxygen react according to the reaction in the video. identify the limiting reacta

GalinKa [24]

Answer:

23.6 g carbon dioxide comes from 8.6 g of CH4 or 10.7 g carbon dioxide comes from 15.6 g O that means the 15.6 g of oxygen is still the limiting reactant because it gets used up and only makes 10.7 g of CO2. 

Explanation:

1) Balanced chemical equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

2) mole ratios:
1 mol CH₄ : 2mol O₂ : 1 mol CO₂ : 2 mol H₂O

3) molar masses
CH₄: 16.04 g/mol
O₂: 32.0 g/mol
CO₂: 44.01 g/mol

4) Convert the reactant masses to number of moles, using the formula

number of moles = mass in grams / molar mass

CH₄: 8.6g / 16.04 g/mol = 0.5362 moles


O₂: 15.6 g / 32.0 g/mol = 0.4875 moles

5) If the whole 0.5632 moles of CH₄ reacted that yields to the same number of moles of CO₂ and that is a mass of:
mass of CO₂ = number of moles x molar mass = 23.60 g of CO₂

Which is what the first part of the answer says.

6) If the whole 0.4875 moles of O₂ reacted that would yield 0.4875 / 2 = 0.24375 moles of CO₂, and that is a mass of:
mass of CO₂ = 0.4875 grams x 44.01 g/mol = 10.7 grams of CO₂.

Which is what the second part of the answer says.

7) From the mole ratio you know infere that 0.5362 moles of CH₄ needs more twice number of moles of O₂, that is 1.0724 moles of O₂, and since there are only 0.4875 moles of O₂, this is the limiting reactant.

Which is what the chosen answer says.

8) From the mole ratios 0.4875 moles of O₂ produce 0.4875 / 2 moles of CO₂, and that is:
0.4875 / 2 mols x 44.01 g/mol = 10.7 g of CO₂, which is the last part of the answer.

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2 years ago

Complete ionic equation K2CO3(aq)+2CuF(aq) → Cu2CO3(s)+2KF(aq) Examine each of the chemical species involved to determine the io

Fudgin [204]

Answer:

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Explanation:

K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)

The complete ionic equation for the above equation can be written as follow:

In solution, K2CO3 and CuF will dissociate as follow:

K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)

CuF(aq) —› Ca^2+(aq) + 2F¯(aq)

Thus, we can write the complete ionic equation for the reaction as shown below:

K2CO3(aq) + 2CuF(aq) —›

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

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2 years ago