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Yuri [45]
1 year ago
5

You have 49.8 g of O2 gas in a container with twice the volume as one with CO2 gas. The pressure and temperature of both contain

ers are the same. Calculate the mass of carbon dioxide gas you have in the container.
Chemistry
1 answer:
IrinaVladis [17]1 year ago
5 0

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

<u>34.2 g is the mass of carbon dioxide gas one have in the container.</u>

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Answer:

false thought ia ion of neon = clarity active

Explanation:

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8 0
2 years ago
Read 2 more answers
Suppose that 0.323 g of an unknown sulfate salt is dissolved in 50 mL of water. The solution is acidified with 6M HCl, heated, a
geniusboy [140]

Answer:

1) 41.16 % = 0.182 grams

2) The alkali cation is K+ , to form the salt K2SO4

Explanation:

Step 1: Data given

Mass of unknown sulfate salt = 0.323 grams

Volume of water = 50 mL

Molarity of HCl = 6M

Step 2: The balanced equation

SO4^2- + BaCl2 → BaSO4 + 2Cl-

Step 3: Calculate amount of SO4^2- in BaSO4

The precipitate will be BaSO4

The amount of SO4^2- in BaSO4 = (Molar mass of SO4^2-/Molar mass BaSO4)*100 %

The amount of SO4^2- in BaSO4 = (96.06 /233.38) * 100

= 41.16%

So in 0.443g of BaSO4 there will be 0.443 * 41.16 % = <u>0.182 grams</u>

<u />

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2. If it is assumed that the salt is an alkali sulfate determine the identity of the alkali cation.

The unknown sulphate salt has 0.182g of sulphate. This means the alkali cation has a weight of 0.323-0.182 = 0.141g grams

An alkali cation has a chargoe of +1; sulphate has a charge of -2

The formula will be X2SO4 (with X = the unknown alkali metal).

Calculate moles of sulphate

Moles sulphate = 0.182 grams (32.1 + 4*16)

Moles sulphate = 0.00189 moles

The moles of sulphate = 0.182/(32.1+16*4)

The moles of sulphate = 0.00189 moles

X2SO4 → 2X+ + SO4^2-

For 2 moles cation we have 1 mol anion

For 0.00189 moles anion, we have 2*0.00189 = 0.00378 moles cation

Calculate molar mass

Molar mass = mass / moles

Molar mass = 0.141 grams / 0.00378 grams

Molar mass = 37.3 g/mol

The closest alkali metal is potassium. (K2SO4 )

3 0
1 year ago
Based on the bond energies for the reaction below, what is the enthalpy of the reaction?HC≡CH (g) + 5/2 O₂ (g) → 2 CO₂ (g) + H₂O
Anuta_ua [19.1K]

Answer:

1219.5 kj/mol

Explanation:

To reach this result, you must use the formula:

ΔHºrxn = Σn * (BE reactant) - Σn * (BE product)

ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H).

The BE values are:

BE C = C: 839 kj / mol

BE C-H: 413 Kj / mol

BE O = O: 495 kj / mol

BE C = O = 799 Kj / mol

BE O-H = 463 kj / mol

Now you must replace the values in the above equation, the result of which will be:

ΔHºrxn = [1 * 839 + 2 * (413) + 5/2 * (495)] - [4 * (799) + 2 * (463) = 1219.5 kj/mol

8 0
1 year ago
Write a balanced half-reaction for the oxidation of liquid water H2O to aqueous hydrogen peroxide H2O2 in basic aqueous solution
nignag [31]

Answer : The balanced half-reaction in a basic solution will be,

2OH^-(aq)\rightarrow H_2O_2(aq)+2e^-  

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Rules for the balanced chemical equation in basic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion (OH^-) at that side where the less number of hydrogen are present.

Now balance the charge.

  • The half reaction is :

H_2O(l)\rightarrow H_2O_2(aq)

  • Now balance the oxygen atoms.

H_2O(l)\rightarrow H_2O_2(aq)+H_2O(l)

  • Now balance the hydrogen atoms.

H_2O(l)+2OH^-(aq)\rightarrow H_2O_2(aq)+H_2O(l)

  • Now balance the charge.

H_2O(l)+2OH^-(aq)\rightarrow H_2O_2(aq)+H_2O(l)+2e^-

The balanced half-reaction in a basic solution will be,

2OH^-(aq)\rightarrow H_2O_2(aq)+2e^-

7 0
2 years ago
How many liters of a 0.0550 M NaF solution contain 0.163 moles of NaF?
Kobotan [32]

Answer : The volume of solution will be 2.96 liters.

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

In this question, the solute is NaF.

Now put all the given values in this formula, we get:

0.0550M=\frac{0.163mole}{\text{Volume of solution (in L)}}

\text{Volume of solution (in L)}=\frac{0.163mole}{0.0550M}

\text{Volume of solution (in L)}=2.96L

Therefore, the volume of solution will be 2.96 liters.

3 0
1 year ago
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