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deff fn [24]
3 years ago
7

At what pressure would 10 moles of Br2 gas have a temperature of 333k In a 77L tank?

Chemistry
1 answer:
Gwar [14]3 years ago
3 0
<h3><u>Answer;</u></h3>

= 3.549 atm

<h3><u>Explanation</u>;</h3>

Using the  PV = nRT

n =number of moles , P is the pressure, V is the volume , T is a constant and R is the Universal constant, 0.08206 L.atm.mol−1.

n = 10 moles

T = 333 K

V = 77 L

P = ?

P × 77 = 0.08206 × 10 × 333

P =  273.2598/77

  <u>= 3.549 atm</u>

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Consider the redox reaction below.
vovangra [49]

Answer:

Zn(s) → Zn⁺²(aq) + 2e⁻

Explanation:

Let us consider the complete redox reaction:

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

This is a redox reaction because, both oxidation and reduction is simultaneously taking place.

  • Oxidation (loss of electrons or increase in the oxidation state of entity)
  • Reduction (gain of electrons or decrease in the oxidation state of the entity)
  • An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.

Here Zn(s) is undergoing oxidation from OS 0 to +2

And H in HCl (aq) is undergoing reduction from OS +1 to 0.

Therefore, for this reaction;

Oxidation Half equation is:

Zn(s) → Zn⁺²(aq) + 2e⁻

Reduction Half equation is:

2H⁺ + 2e⁻ → H₂(g)

4 0
3 years ago
How many grams of acetylene are produced by adding water to 5.00 g CaC2?
Murljashka [212]

Answer:

2.03125g of acetylene

Explanation:

First thing's first, we have to write out the balanced chemical equation;

CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g)

Water is in excess, so CAC2 is our limiting reactant. i.e it determines the amount of product that would be formed.

1 mol of CaC2 produces 1 mol of C2H2

In terms of mass;

Mass = Number of moles * Molar mass

where the molar mass of the elements are;

Ca = 40g/mol

C = 12g/mol

H = 1g/mol

CaC2 = 40+ (2*12) = 64g/mol

C2H2 =( 2 * 12) + ( 2 * 1) = 26g/mol

64g (1 * 64g/mol) of CaC2 produces 26g ( 1mol * 26g/mol) of C2H2

5g would produce x?

64 = 26

5 = x

Upon solving for x we have;

x = (5 * 26) / 64

x = 2.03125g

5 0
3 years ago
please hurry! If 3.87g of powdered aluminum oxide is placed in a container containing 5.67g of water, what is the limiting react
VikaD [51]

Answer:

Explanation:

Given parameters:

Mass of aluminium oxide = 3.87g

Mass of water = 5.67g

Unknown:

Limiting reactant = ?

Solution:

The limiting reactant is the reactant in short supply in a chemical reaction. We need to first write the chemical equation and convert the masses given to the number of moles.

Using the number of moles, we can ascertain the limiting reactants;

         Al₂O₃  + 3H₂O  →  2Al(OH)₃  

Number of moles;

            Number of moles = \frac{mass}{molar mass}

molar mass of Al₂O₃  = (2x27) + 3(16) = 102g/mole

    number of moles = \frac{3.87}{102}  = 0.04mole

   

molar mass of  H₂O = 2(1) + 16 = 18g/mole

    number of moles = \frac{5.67}{18}  = 0.32mole

From the reaction equation;

        1 mole of  Al₂O₃  reacted with 3 moles of H₂O

   0.04 mole of Al₂O₃ will react with 3 x 0.04 mole = 0.12 mole of H₂O

But we were given 0.32 mole of H₂O and this is in excess of amount required.

This shows that Al₂O₃ is the limiting reactant

           

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