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ikadub [295]
3 years ago
11

If 3.0 liters of oxygen gas react with excess carbon monoxide at STP, how many liters of carbon dioxide can be produced under th

e same conditions?2 CO (g) + O2 (g) ---> 2 CO2 (g)
a. 1.5 L
b. 3.0 L
c. 4.5 L
d. 6.0 L,
Chemistry
1 answer:
mr_godi [17]3 years ago
8 0

Answer:

The volume of CO2 produced is 6.0 L (option D)

Explanation:

Step 1: Data given

Volume of oxygen = 3.0 L

Carbon monoxide = CO = in excess

Step 2: The balanced equation

2 CO (g) + O2 (g) → 2 CO2 (g)

Step 3: Calculate moles of O2

1 mol of gas at STP = 22.4 L

3.0 L = 0.134 moles

Step 3: Calculate moles of CO2

For 2 moles CO we need 1 mol of O2 to produce 2 moles of CO2

For 0.134 moles O2 we'll have 2*0.134 = 0.268 moles CO2

Step 4: Calculate volume of CO2

1 mol = 22.4 L

0.268 mol = 22.4 * 0.268 = 6.0 L

The volume of CO2 produced is 6.0 L

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What is the mass number of an element with an atomic number 10 and 10 neutrons
Bogdan [553]

Answer:

Neon

Explanation:

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3 years ago
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Matter and energy may be created or destroyed in a chemical reaction.
gulaghasi [49]

Answer:false

Explanation:matter cannot be created or destroyed in a chemical reaction it is rearranged

3 0
3 years ago
What is the overall enthalpy of reaction for the equation shown below?
Rudiy27

Answer:

ΔH₁₂ = -867.2 Kj

Explanation:

Find enthalpy for 3H₂ + O₃ => 3H₂O given ...

2H₂ + O₂ => 2H₂O      ΔH₁ = -483.6 Kj

        3O₂ => 2O₃        ΔH₂ = + 284.6 Kj

_____________________________

3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O       (multiply by 3 to cancel O₂)

6H₂ + 3O₂ => 6H₂O        ΔH₁ = 3(-483.6 Kj) = -1450.6Kj

          2O₃ => 3O₂           ΔH₂ = -284.6Kj              (reverse rxn to cancel O₂)

_______________________________

6H₂ + 2O₃ => 6H₂O         ΔH₁₂ = -1735.2 Kj       (Net Reaction - not reduced)

________________________________

divide by 2 => target equation (Net Reaction - reduced)

3H₂ + O₃ => 3H₂O            ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj    

4 0
3 years ago
A 0.533 g sample of a carbonate salt is added to a flask. The only thing you know is that per formula unit, there is only one ca
dimulka [17.4K]

Answer:

\mathbf{CrCO_3}

Explanation:

The total mole of H_2SO_4 added = 15 × 1 × 10⁻³

= 15 × 10⁻³ mole

= 15 mmoles

the number of moles of NaOH added in order to neutralize the excess acid = 0.5905 ×  29.393 = 17.36 mmoles

the equation for the reaction is:

2NaOH + H_2SO_4 --------> Na_2SO_4 +2H_2O

i.e

2 moles of NaOH react with H_2SO_4

1 moles of  NaOH will react with 1/2  H_2SO_4

17.36 mmoles of NaOH = 1/2 × 17.36 mmoles of H_2SO_4

                                       = 8.68 mmoles

Number of moles of H_2SO_4 that react with MCO₃ = Total moles of H_2SO_4 added - moles of H_2SO_4 reacted with NaOH

= (15 - 8.68) mmoles

= 6.32 mmoles

H_2SO_4 + MCO_3  ------>   M_2SO_4   +    H_2O   +   CO_2

1 mole of H_2SO_4 react with 1 mole of MCO_3

6.32 mmoles of H_2SO_4 = 6.32 mmoles of MCO_3

number of moles of MCO_3 = 6.32 10⁻³ moles

mass of  MCO_3 (carbonate salt) = 0.533 g

molar mass of MCO_3  = (M+60)g/mol

We all know that ;

number of moles = mass/molar mass

Then:

6.32 10⁻³ = 0.533 / (M+60)

(M+60) = 0.533/ 6.32 10⁻³

M + 60 = 84.34

M = 84.34 - 60

M = 24.34

Thus the element with the atomic mass of 24.34 is Chromium

The chemical formula for the compound is :  \mathbf{CrCO_3}

6 0
3 years ago
Number 8!!!!! pls I don't get it
vlada-n [284]

I think the answer would be 1.58 g.

5 0
3 years ago
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