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pochemuha
3 years ago
10

What is the pressure exerted by 68.0 g of nitrogen trihydride gas in a 50.0L container at 30.0 C?

Chemistry
1 answer:
Cloud [144]3 years ago
8 0

Answer:

Molecular mass of NH3 = ( 14 + 3 ) = 17 g

PV =  \frac{m}{m _{r}} RT \\ P \times 50.0 \times  {10}^{ - 6}  =  \frac{68.0}{17}  \times 8.314 \times (30.0 + 273) \\ P =  \frac{68.0 \times 8.314 \times 303}{17 \times 50.0 \times  {10}^{ - 6} }  \\ P = 2.02 \times  {10}^{8}  \: Pascals

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A Calorie unit used in food is equal to the amount of energy necessary to raise the temperature of 1 kilogram of water ________
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7 0
2 years ago
Which is more metallic<br><br> a) Silicon<br><br> b) Aluminum
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4 0
3 years ago
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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes
frez [133]

Answer : The mass of MnCO_3 required are, 35 kg

Explanation :

First we have to calculate the mass of MnO_2.

The first step balanced chemical reaction is:

2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2

Molar mass of MnCO_3 = 115 g/mole

Molar mass of MnO_2 = 87 g/mole

Let the mass of MnCO_3 be, 'x' grams.

From the balanced reaction, we conclude that

As, (2\times 115)g of MnCO_3 react to give (2\times 87)g of MnO_2

So, xg of MnCO_3 react to give \frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg of MnO_2

And as we are given that the yield produced from the first step is, 65 % that means,

60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg

The mass of MnO_2 obtained = 0.4542x g

Now we have to calculate the mass of Mn.

The second step balanced chemical reaction is:

3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3

Molar mass of MnO_2 = 87 g/mole

Molar mass of Mn = 55 g/mole

From the balanced reaction, we conclude that

As, (3\times 87)g of MnO_2 react to give (3\times 55)g of Mn

So, 0.4542xg of MnO_2 react to give \frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg of Mn

And as we are given that the yield produced from the second step is, 80 % that means,

80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg

The mass of Mn obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g     (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of MnCO_3 required are, 35 kg

4 0
3 years ago
Read 2 more answers
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