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3 years ago

How many moles are there in 78.3g of CO2?

1 answer:

4 0

The molar mass of CO2 can be calculated as follows;
CO2 — 12 + (16x2) = 12+ 32 = 44 g
Therefore molar mass of CO2 is 44 g/mol
In 44 g of CO2 there’s 1 mol of CO2
Then 1 g of CO2 there’s 1/44 mol of CO2
Therefore in 78.3 g of CO2 there’s — 1/44 x 78.3 =1.78 mol of CO2

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What is the vapor pressure (in mm Hg) of a solution of 17.5 g of glucose (C6H12O6) in 82.0 g of methanol (CH3OH) at 27∘C? The va

vichka [17]

Answer:

134.8 mmHg is the vapor pressure for solution

Explanation:

We must apply the colligative property of lowering vapor pressure, which formula is: P° - P' = P° . Xm

P° → Vapor pressure of pure solvent

P' → Vapor pressure of solution

Xm → Mole fraction for solute

Let's determine the moles of solute and solvent

17.5 g . 1 mol/180 g = 0.0972 moles

82 g . 1mol / 32 g = 2.56 moles

Total moles → moles of solute + moles of solvent → 2.56 + 0.0972 = 2.6572 moles

Xm → moles of solute / total moles = 0.0972 / 2.6572 = 0.0365

We replace the data in the formula

140 mmHg - P' = 140 mmHg . 0.0365

P' = - (140 mmHg . 0.0365 - 140mmHg)

P' = 134.8 mmHg

5 0

3 years ago

Which of the following samples will have the greatest volume at STP?

Tamiku [17]

Answer:

Option b. 22 g of He will have the greatest volume at STP

Explanation:

In order to determine the volume, we apply the Ideal Gases Law equation:

P . V = n . R . T

V = n . R . T / P

R, T and P are the same in all the situation we must define n (number of moles).

The one that has the greatest number of moles will have the greatest volume at STP

22 g of Ne . 1mol / 20.1 g = 1.09 moles of Ne

22g of He . 1mol / 4 g = 5.5 moles of He

22 g of O₂ . 1mol / 32g = 0.68 moles of O₂

22 g of Cl₂ . 1mol / 70.9 g = 0.31 moles of Cl₂

4 0

2 years ago

Sodium peroxide (Na2O2) is often used in self-contained breathing devices, such as those used in fire emergencies, because it re

Anastaziya [24]

Answer:

725.15 L

Explanation:

The balanced chemical equation for the reaction between Na₂O₂ and CO₂ is the following:

Na₂O₂ + CO₂ → Na₂CO₃ + 1/2 O₂

From the stoichiometry of the reaction, 1 mol of Na₂O₂ reacts with 1 mol CO₂. So, the stoichiometric ratio is 1 mol Na₂O₂/1 CO₂.

Now, we convert the mass of reactants to moles by using the molecular weight (Mw) of each compound:

Mw (Na₂O₂) = (23 g/mol x 2) + (16 g/mol x 2) = 78 g/mol

moles Na₂O₂ = 96.7 g/(78 g/mol) = 1.24 mol Na₂O₂

Mw (CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol

moles CO₂ = 0.0755 g/(44 g/mol) = 1.71 x 10⁻³ mol CO₂

Now, we calculate the number of moles of CO₂ we need to completely react with the mass of Na₂O₂ we have:

1.24 mol Na₂O₂ x 1 mol CO₂/1 mol Na₂O₂ = 1.24 mol CO₂

In 1 L of respired air we have 1.71 x 10⁻³ mol CO₂ (0.0755 g), so we need the following number of liters to have 1.24 mol of CO₂:

1.24 mol CO₂ x 1 L/1.71 x 10⁻³ mol CO₂ = 725.15 L

4 0

3 years ago

A 2.684-g sample of zinc oxide was reduced by hydrogen gas, resulting in 2. 156 g of pure zinc metal. Determine the empirical fo

Afina-wow [57]

The empirical formula of the initial zinc oxide is ZnO.

<h3>What is Empirical Formula?</h3>

The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.

It is the lowest whole number ratio of the element in the compound.

<h3>How to find out the empirical formula?</h3>

Find out the given masses and molar masses of the elements

The molar mass of Zn = 65 gmol⁻¹

Given the mass of Zn = 2.156 g

The molar mass of Oxygen = 16 gmol⁻¹

The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal

= (2.684 - 2.156) g

= 0.528 g

Find the number of moles of the elements in the compound

The number of moles is given by

$n = \frac{m}{M}$

where m = given mass and

M = Molar mass

Number of moles of Zinc = $\frac{2.156}{65}$ = 0.033 moles

Number of moles of Oxygen = $\frac{0.528}{16}$ = 0.033 moles

Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.

Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.

Therefore, the empirical formula of zinc oxide is ZnO.

Learn more about the empirical formula:

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5 0

1 year ago

What is the difference between an Isotope and an Allotrope?

Kaylis [27]

Allotropes are elements on the periodic table that have more than one crystalline form . There are three forms of the element carbon : Diamond, Graphite, and Fullerenes. Isotopes are atoms of the same element with the same atomic number but have a different mass number.

8 0

3 years ago