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Svetradugi [14.3K]
2 years ago
15

4NH3+5O2-4NO+6H2Ohow many moles of NH3 must react to produce 5.0 moles of NO?​

Chemistry
1 answer:
tresset_1 [31]2 years ago
7 0

Answer: 5.0 moles

Explanation:

From the equation, we see that for every 4 moles of ammonia consumed, 4 moles of nitrogen monoxide are produced (we can reduce this to moles of ammonia consumed = moles of nitrogen monoxide produced).

This means that the answer is <u>5.0 mol</u>

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Calculate the pressure of 2 mol of a gas at 300 K in 8 L container.
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2*300*8=4800

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The proximate reason for the uptake by plants of nutrients like ammonium nitrate is?
Blizzard [7]

The proximate reason for the uptake by plants of nutrients like ammonium nitrate is homeostatis

<h3>What is homeostasis?</h3>

Homeostasis refers to an organism's ability to regulate various physiological processes to keep internal states steady and balanced.

<h3>How does plant perform homeostatis?</h3>

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Much like when we sweat, this allows the plant to cool and maintain homeostasis.

Similarly, the same process allows plants to  absorb nutrient like ammonium nitrate.

Thus, the proximate reason for the uptake by plants of nutrients like ammonium nitrate is homeostatis.

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1 year ago
Classify each of the following properties by writing physical or chemical matter
Ivenika [448]
1. Physical matter
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6 0
3 years ago
How many milliliters of 0.125 M FeCl3 are needed to react with an excess of Na2S to produce 3.75 g of Fe2S3 if the percent yield
Katyanochek1 [597]

Answer:

0.912 mL

Explanation:

3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)

FeCl3 is the limiting reactant.

Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles

Hence actual yield of Iron III sulphide = 0.043 moles

Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield

Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide

From the reaction equation,

2moles of iron III chloride produced 1 mole of iron III sulphide

x moles of iron III chloride, will produce 0.057 of iron III sulphide

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Volume= number of moles/ concentration

Volume= 0.114/0.125

Volume= 0.912 mL

4 0
3 years ago
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