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3 years ago

Hydrogen gas can combine with oxygen gas under certain conditions to form water.

Chemistry

1 answer:

stellarik [79]3 years ago

5 0

Answer:

The correct answer is product.

Explanation:

Hydrogen gas reacts with oxygen gas to generate water molecule.

2H2+O2 = 2H2O

We all know that that substances which are present at the left hand side of the equilibrium sign act as substrate and the substances those are present in the right hand side of the equilibrium sign are called products.

From that point of view it can be stated that water or H2O is the product of the reaction between hydrogen and oxygen.

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For a particular reaction, ΔH=−111.4 kJ/mol and ΔS=−25.0 J/(mol·K).

PtichkaEL [24]

That’s the answer ^^^^

3 0

3 years ago

What is the mass of 5 mole of ammonia . Calculate the number of NH₃ molecules, nitrogen atom and hydrogen atoms in it..

Aloiza [94]

Molar mass of NH_3

$\\ \sf\longmapsto 14u+3(1u)$

$\\ \sf\longmapsto 14u+3u$

$\\ \sf\longmapsto 17g/mol$

We know.

No of moles=Given mass/Molar mass

$\\ \sf\longmapsto Given\;Mass=17(5)$

$\\ \sf\longmapsto Given \:Mass\:of\:NH_3=85g$

Now

Lets write the balanced equation

$\\ \sf\longmapsto N_2+3H_2=2NH_3$

There is 2moles of Ammonia
3moles of H_2
1mole of N_2

Now

$\boxed{\sf No\:of\:Molecules =No\:of\;moles\times Avagadro\:no}$

For Hydrogen

$\\ \sf\longmapsto 3\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 18.069\times 10^{23}$

$\\ \sf\longmapsto 1.8\times 10^{22}molecules$

For Ammonia

$\\ \sf\longmapsto 2\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 12.046\times 10^{23}$

$\\ \sf\longmapsto 1.2\times 10^{22}molecules$

For Nitrogen

$\\ \sf\longmapsto 1\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 6.023\times 10^{23}molecules$

3 0

3 years ago

You’ve just started a new job in the laboratory of the Food and Drug Administration. Your supervisor asks you to prepare a 1.50-

anastassius [24]

Answer: 118.5 grams

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute

$V_s$ = volume of solution in ml = 500 ml

$1.50=\frac{moles\times 1000}{500ml}$

moles = 0.75

moles of solute = $\frac{\text {given mass}}{\text {molar mass}}$

0.75 = $\frac{\text {given mass}}{158g/mol}$

mass of $KMnO_4$ = 118.5 grams

Thus mass of $KMnO_4$ needed to prepare 500 mL of this solution iis 118.5 grams

6 0

3 years ago

Where two air masses and do not mix

yulyashka [42]

When two air masses meet together, the boundary between the two is called a weather front. At a front, the two air masses have different densities, based on temperature, and do not easily mix. One air mass is lifted above the other, creating a low pressure zone.

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3 years ago

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zloy xaker [14]

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3 years ago