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Usimov [2.4K]
3 years ago
8

10. If 200. L of ammonia (NH3) are produced in the reaction between nitrogen and hydrogen at STP, how many

Chemistry
1 answer:
erma4kov [3.2K]3 years ago
8 0

Answer:

100L of N2

Explanation:

First let us obtain the number of mole of NH3 that occupied 200L at stp. This can be achieved by doing the following:

1mole of NH3 occupied 22.4L at stp.

Therefore, Xmol mole of NH3 will occupy 200L i.e

Xmol of NH3 = 200/22.4 = 8.93moles.

Now let us generate a balanced equation for the reaction.

N2 + 3H2 —> 2NH3

From the equation above,

1mole of N2 produced 2moles of NH3.

Therefore, Xmol of N2 will produce 8.93moles of NH3 i.e

Xmol of N2 = 8.93/2 = 4.47moles

Now let us convert 4.47moles to N2 to Litres. This is illustrated below:

1mole of N2 occupied 22.4L at stp

Therefore, 4.47moles of N2 will occupy = 4.47 x 22.4 = 100L

Therefore, 100L of N2 is used in the reaction

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3 years ago
Calculate the final Celsius temperature when 634 L at 21 °C is compressed to 307 L.
Illusion [34]

Answer:

- 130.64°C.

Explanation:

  • We can use the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n and P are constant, and have two different values of V and T:

<em>V₁T₂ = V₂T₁</em>

<em></em>

V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.

V₂ = 307.0 L, T₂ = ??? K.

<em>∴ T₂ = V₂T₁/V₁ </em>= (307.0 L)(294.0 K)/(634.0 L) = <em>142.36 K.</em>

<em>∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.</em>

5 0
3 years ago
A student heats 2.796g of zinc powder with 2.414g of sulphur.He reported that he obtained 4.169g of zinc sulfide and recover 1.0
Sati [7]

Answer:

The mass of reactants and products are equal hence the reaction obeys law of conservation of mass

Explanation:

The law of mass conservation states that for a closed system to all transfer of mass, the mass of system must remain constant over time. This means for a chemical reaction, the mass of reactants must equal the mass of products.

if 2.796g of Zn reacts with 2.414g of sulphur to produce 4.169g of ZnS ad 1.041g of unreacted sulphur, then it means that accorfing to the law of mass conservation, the mass of reactants (zinc and sulphur), must be equal to mass of products (zinc sulfide and unreacted sulphur)

Mass of reactants = 2.796g + 2.414g =5.21g

Mass of products = 4.169g + 1.041g=5.21g

5 0
3 years ago
If my aunt means degree Celsius or degrees Fahrenheit
JulijaS [17]
If it is 60 Celsius that would conver to fare height by means of this equation; (1.8*60)+32°F
Which would come out to.... 140° Fahrenheit... Hardly seems like chilly conditions.
7 0
3 years ago
How many milliliters of an aqueous solution of 0.170 M ammonium carbonate is needed to obtain 16.1 grams of the salt
Citrus2011 [14]

There will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.There will be needed mL of

Why?

In order to calculate how many milliliters are needed to obtain 16.1 grams of the salt given its concentration, we first need to find its chemical formula which is the following:

(NH_{4})2CO_{3}

Now that we know the chemical formula of the substance, we need to find its molecular mass. We can do it by the following way:

N_{2}=14g*2=28g\\\\2H_{4}=2*1g*4=8g\\\\C=12.01g*1=12.01g\\\\O_{3}=15.99g*3=47.97g

We have that the molecular mass of the substance will be:

MolecularMass=\frac{28g+8g+12.01g+47.97g}{mol}=95.98\frac{g}{mol}

Therefore, knowing the molecular mass of the substance, we need to calculate how many mols represents 16.1 grams of the same substance, we can do it by the following way:

mol_{(NH_{4})2CO_{3}=\frac{mass_{(NH_{4})2CO_{3}}}{molarmass_{(NH_{4})2CO_{3}}}

mol_{(NH_{4})2CO_{3}=\frac{16.1g}{95.98\frac{g}{mol}}=0.167mol

Finally, if we need to calculate how many milliliters are needed, we need to use the following formula:

M=\frac{moles_{solute}}{volume_{solution}}

M=\frac{moles_{solute}}{volume_{solution}}\\\\volume_{solution}=\frac{moles_{solute}}{M}

Now, substituting and calculating, we have:

volume_{solution}=\frac{0.167mol}{0.170\frac{mol}{L}}\\\\volume_{solution}=0.982L=0.982L*1000=982.35mL

Henc, there will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.

Have a nice day!

5 0
3 years ago
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