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defon
2 years ago
5

What is the general form of a decomposition reaction

Chemistry
1 answer:
likoan [24]2 years ago
8 0

A reaction where one becomes two


2H202 --> 2H2O + O2

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Germanium is a group 4A semiconductor. The addition of a dopant atom (group 3A element) that has fewer valence electrons than th
klasskru [66]

Answer:

Doping with galium or indium will yield a p-type semiconductor while doping with arsenic, antimony or phosphorus will yield an n-type semiconductor.

Explanation:

Doping refers to improving the conductivity of a semiconductor by addition of impurities. A trivalent impurity leads to p-type semiconductor while a pentavalent impurity leads to an n-type semiconductor.

7 0
2 years ago
If a scientist has a 105 mL solution of 0.08 M HCl, then to a good approximation, what will be the pH of the solution? Enter you
djyliett [7]

Answer:

5667

Explanation:

6 0
2 years ago
2N2H4+ N2O4———3N2+4H2O SalmaKhan99 avatar How many grams of N2 gas will be formed by reacting 100g of N2H4 and 200g of N2 Kindly
Alona [7]

Answer:

131.26 g

Explanation:

From the balanced equation,

2 moles of N₂H₄reacts with 1 mole of N₂O₄ to give 3 moles of N₂

Now number of moles of N₂H₄ present in 100 g N₂H₄ is n = 100 g/molar mass N₂H₄.

Molar mass N₂H₄ = 2 × 14.01 g/mol + 1 × 4 g/mol = 28.02 g/mol + 4 g/mol = 32.02 g/mol

n₁ = 100/32.02 = 3.123 mol

Also

Now number of moles of N₂O₄ present in 200 g N₂O₄ is n = 200 g/molar mass N₂O₄.

Molar mass N₂O₄ = 2 × 14.01 g/mol + 16 × 4 g/mol = 28.02 g/mol + 64 g/mol = 92.02 g/mol

n₂ = 200/92.02 = 2.173 mol

Since the mole ratio of N₂H₄  to N₂O₄ is 2 : 1, We require 2 × 2.173 mol N₂H₄  to react with 2.173 mole N₂O₄  

Number of moles of N₂H₄ required is 4.346. But the number of moles of N₂H₄  present is 3.123 so N₂H₄  is the limiting reagent.

So, from the equation, 2 moles of N₂H₄ produces 3 moles of N₂

Therefore number of mole N₂ = 3/2 moles of N₂H₄ = 3/2 × 3.123 mol = 4.6845 mol

From n = m/M where n = number of moles of nitrogen gas = 4.6845 mol and M = molar mass of nitrogen gas = 28.02 g/mol and m = mass of nitogen gas.

m = nM = 4.6845 mol × 28.02 g/mol = 131.26 g

So the mas of nitrogen gas produced is 131.26 g

4 0
3 years ago
A first-order decomposition reaction has a rate constant of 0.00440 yr−1. How long does it take for [reactant] to reach 12.5% of
mina [271]

Answer:

473 year

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given:

To reach 12.5% of reactant means that 0.125 of [A_0] is decomposed. So,

\frac {[A_t]}{[A_0]} = 0.125

t = ?

\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.125=e^{-0.00440\times t}

t = 473 year

8 0
3 years ago
Balloon has a volume of 600-ml at temperature of 360 K. If the temperature of
Molodets [167]

Answer:

V₂ ≈416.7 mL

Explanation:

This question asks us to find the volume, given another volume and 2 temperatures in Kelvin. Based on this information, we must be using Charles's Law and the formula. Remember, his law states the volume of a gas is proportional to the temperature.

  • V₁ / T₁ = V₂ / T₂

where V₁ and V₂ are the first and second volumes, and T₁ and T₂ are the first and second temperature.

The balloon has a volume of 600 milliliters and a temperature of 360 K, but the temperature then drops to 250 K. So,

  • V₁= 600 mL
  • T₁= 360 K
  • T₂= 250 K

Substitute the values into the formula.

  • 600 mL /360 K = V₂ / 250 K

Since we are solving for the second volume when the temperature is 250 K, we have to isolate the variable V₂. It is being divided by 250 K. The inverse o division is multiplication, so we multiply both sides by 250 K.

  • 250 K * 600 mL /360 K = V₂ / 250 K * 250 K
  • 250 K * 600 mL/360 K = V₂

The units of Kelvin cancel, so we are left with the units of mL.

  • 250 * 600 mL/360=V₂
  • 416.666666667 mL= V₂

Let's round to the nearest tenth. The 6 in the hundredth place tells us to round to 6 to a 7.

  • 416.7 mL ≈V₂

The volume of the balloon at 250 K is approximately 416.7 milliliters.

5 0
2 years ago
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