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Aneli [31]
3 years ago
14

The decomposition of ammonia to nitrogen and hydrogen on a tungsten filament at 800°C is independent of the concentration of amm

onia at high pressures of ammonia. What is the order of the reaction with respect to ammonia?
Chemistry
1 answer:
finlep [7]3 years ago
7 0

Answer:

Order zero

Explanation:

Let's consider the decomposition of ammonia to nitrogen and hydrogen on a tungsten filament at 800°C.

2 NH₃(g) → N₂(g) + 3 H₂(g)

The generic rate law is:

rate = k × [NH₃]ⁿ

where,

rate: reaction rate

k: rate constant

n: reaction order

When n = 0, we get:

rate = k × [NH₃]⁰ = k

As we can see, when the reaction order with respect to ammonia is zero, the reaction rate is independent of the concentration of ammonia.

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The first-order decay of radon has a half-life of 3.823 days. How many grams of radon remain after 7.22 days if the sample initi
Nesterboy [21]

substitute: <span><span>t<span>1/2</span></span>=<span><span>ln(2)</span>k</span>→k=<span><span>ln(2)</span><span>t<span>1/2</span></span></span></span> Into the appropriate equation: <span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−kt</span></span></span> <span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−<span><span>ln(2)</span><span>t<span>1/2</span></span></span>t</span></span></span> <span>[A<span>]t</span>=(250.0 g)∗<span>e<span>−<span><span>ln(2)</span><span>3.823 days</span></span>(7.22 days)</span></span>=67.52 g</span>

6 0
3 years ago
If a gas is initially at a pressure of 9 atm, a volume of 21 liters, and a temperature of 253 K, and then the pressure is raised
alex41 [277]

Answer:

15.04 mL

Explanation:

Using Ideal gas equation for same mole of gas as

\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}

Given ,  

V₁ = 21 L

V₂ = ?

P₁ = 9 atm

P₂ = 15 atm

T₁ = 253 K

T₂ = 302 K

Using above equation as:

\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}

\frac {{9}\times {21}}{253}=\frac {{15}\times {V_2}}{302}

Solving for V₂ , we get:

<u>V₂ = 15.04 mL</u>

3 0
2 years ago
A chemist has to prepare 250.0 mL of a 0.300 M Na2SO4(aq) solution. What mass, in grams, of sodium sulfate (formula mass 142.05
Sedbober [7]

The mass of sodium sulphate, Na₂SO₄, required to prepare the solution is 10.65 g

<h3>How to determine the mole of sodium sulphate Na₂SO₄</h3>
  • Volume = 250 mL = 250 / 1000 = 0.25 L
  • Molarity = 0.3 M
  • Mole of Na₂SO₄ =?

Mole = Molarity x Volume

Mole of Na₂SO₄ = 0.3 × 0.25

Mole of Na₂SO₄ = 0.075 mole

<h3>How to determine the mass of sodium sulphate Na₂SO₄</h3>
  • Molar mass of Na₂SO₄ = 142.05 g/mol
  • Mole of Na₂SO₄ = 0.075 mole
  • Mass of Na₂SO₄ =?

Mass = mole × molar mass

Mass of Na₂SO₄ = 0.075 × 142.05

Mass of Na₂SO₄ = 10.65 g

Thus, 10.65 g of Na₂SO₄ is needed to prepare the solution.

Learn more about molarity:

brainly.com/question/15370276

6 0
1 year ago
PLEASE HELPPPP!!!!!!!!!!!!
bixtya [17]

Answer:

Largest: Mg

Smallest: Cl

Explanation:

8 0
3 years ago
Is this atom more likely to gain electrons or to lose electrons? Explain how you can tell
elena55 [62]

Answer:

Likely to gain electrons

Explanation:

The atom shown is likely to gain additional electrons to complete its electronic configuration.

  • Since this is a neutral specie, the number of protons and electrons are the same.
  • The atom has 16 electrons
  • the number of valence electrons is 6
  • If the atom gains two additional electrons, the octet configuration is attained
  • Also, the atom can lose 6 electrons to become an octet

The atom will prefer to gain additional 2 electrons to give an octet configuration.

5 0
2 years ago
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