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Bezzdna [24]
3 years ago
6

Refer to the following balanced equation in which ammonia reacts with nitrogen monoxide to produce nitrogen and water.

Chemistry
1 answer:
RSB [31]3 years ago
6 0

Answer:4314

Explanation:

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In a chemical reaction, one of the chemical reactants is present in______ to make sure that the limiting reagent is completely c
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Trace amounts of sulfur (S) in coal are burned in the presence of diatomic oxygen (O2) to form sulfur dioxide (SO2). Determine t
Mashcka [7]

Answer:

0.99 kg O₂

1.9 kg SO₂

Explanation:

Let's consider the reaction between sulfur and oxygen to form sulfur dioxide.

S + O₂ → SO₂

The mass ratio of S to O₂ is 32.07:32.00. The mass of oxygen required to react with 1 kg of sulfur is:

1 kg S × (32.00 kg O₂/32.07 kg S) = 0.998 kg O₂

The mass ratio of S to SO₂ is 32.07:64.07. The mass of sulfur dioxide formed when 1 kg of sulfur is burned is:

1 kg S × (64.07 kg SO₂/32.07 kg S) = 1.99 kg SO₂

3 0
3 years ago
What type of ion does the substance give off if it turns the pH paper blue?
olga_2 [115]

Answer:

Alkaline

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In an alkaline solution, red litmus paper turns blue. When an alkaline compound dissolves in water, it produces hydroxide ions, which cause the solution to become alkaline.

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For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coeffici
Brut [27]

<u>Question:</u>

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K

<u>Answer:</u>

The mean activity coefficient for HCl solution is 0.78.

<u>Explanation:</u>

Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is

              E=E_{0}-0.0514 \mathrm{V} \ln \gamma

As we know that E_{0} = 0.22 V and E = 0.342 V, the equation will become

             0.342 V+0.0514 V \ln (0.124)=0.22 V-0.0514 V \ln \gamma

             0.342 V-0.222 V=-0.0514 V(\ln \gamma+\ln 0.124)

             0.12 \mathrm{V}=-0.0514 \mathrm{V}(\ln \gamma+\ln 0.124)

             \frac{0.12}{0.0514}=-\ln (0.124 \gamma)

             -2.3346=\ln (0.124 \gamma)

             e^{-2.3346}=0.124 \gamma

             0.0968=0.124 \gamma

             \gamma=\frac{0.0968}{0.124}=0.78

So, the mean activity coefficient is 0.78.

6 0
3 years ago
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