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Trava [24]
3 years ago
9

How many grams of sodium sulfide can be produced when 45.3 g Na react with 105 g S?

Chemistry
1 answer:
erastova [34]3 years ago
8 0
Balance the equation: 2Na + S --> Na2S
Using the given amount of the reactants in the reaction, calculate the amount of the product:
45.3g Na x (1 mol/22.99 g)= 1.97 mol of Na
105f S x (1 mol/ 32.06g) = 3.28 mol of S
The limiting reactant would be Na:
<span>1.97 mol Na x (1 mol Na2S/ 2 mol Na) x (78.04g/mol) = 76.87g of Na2S produced</span>
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4.45 kcal of heat was added to increase the temperature of a sample of water from 23.0 °C to 57.8 °C. Calculate
Alona [7]

Answer:

m = 4450 g

Explanation:

Given data:

Amount of heat added = 4.45 Kcal ( 4.45 kcal ×1000 cal/ 1kcal = 4450 cal)

Initial temperature = 23.0°C

Final temperature = 57.8°C

Specific heat capacity of water = 1 cal/g.°C

Mass of water in gram = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 57.8°C - 23.0°C

ΔT = 34.8°C

4450 cal = m × 1 cal/g.°C × 34.8°C

m = 4450 cal / 1 cal/g

m = 4450 g

4 0
3 years ago
Which statement is true regarding a hydrogen bond? It is weaker than dipole interaction forces. It is weaker than London dispers
Rom4ik [11]
I believe the answer is C: "<span>It occurs when a hydrogen atom bonds with electropositive atoms."</span>
4 0
3 years ago
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Number 2 &amp; 3 please !!&lt;3
Goshia [24]

Answer:just for the ponits

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8 0
2 years ago
What property do atoms of these elements have that helps make the molecules polar
DENIUS [597]
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4 0
3 years ago
A mixture of 9.22 moles of A, 10.11 moles of B, and 27.83 moles of C is placed in a one-liter
AysviL [449]

The equilibrium constant is 1.3 considering the reaction as written in the question.

<h3>Equilibrium in chemical reactions</h3>

In a chemical reaction, the equilibrium constant is calculated based on the equilibrium concentration of each specie. The equation of this reaction is;

A (g) + 2B (g) ⇌ 3C (g).

The initial concentration of each specie is;

  • A - 9.22 M
  • B - 10.11 M
  • C - 27.83 M

The equilibrium concentration of B is 18.32 M

We now have to set up the ICE table as follows;

         A (g) +    2B (g) ⇌     3C (g)

I       9.22        10.11           27.83

C     -x              -x                +x

E   9.22 - x       10.11 - x       27.83 + x

The equilibrium concentration of B is 18.32 M hence;

10.11 - x = 18.32

x = 10.11 - 18.32 = -8.21

Hence;

Equilibrium concentration of A = 9.22 - (-8.21) = 17.43

Equilibrium concentration of C = 27.83 + (-8.21) = 19.62

Equilibrium constant K = [19.62]^3/[17.43] [18.32]^2

K = 1.3

Learn more about equilibrium constant: brainly.com/question/17960050

4 0
2 years ago
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