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KatRina [158]
3 years ago
10

What is the molar mass of Ca(NO3)2?

Chemistry
1 answer:
gregori [183]3 years ago
7 0

Answer:

C.164.1g/mol

Explanation:

40+14×2+16×6

164

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A sample of gas occupies 300 mL at 100K. What is its volume when the
iris [78.8K]

If the temperature of the sample of gas increases to the given value, the volume also increases to 600mL.

<h3>What is Charles's law?</h3>

Charles's law states that "the volume occupied by a definite quantity of gas is directly proportional to its absolute temperature.

It is expressed as;

V₁/T₁ = V₂/T₂

Given the data in the question;

  • Initial temperature of gas T₁ = 100K
  • Initial volume of gas V₁ = 300mL
  • Final temperature T₂ = 200K
  • Final volume V₂ = ?

V₁/T₁ = V₂/T₂

V₂ = V₁T₂ / T₁

V₂ = ( 300mL × 200K ) / 100K

V₂ = 60000mLK / 100K

V₂ = 600mL

Therefore, if the temperature of the sample of gas increases to the given value, the volume also increases to 600mL.

Learn more about Charles's law here: brainly.com/question/12835309

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3 0
2 years ago
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
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What is the formula of the ionic compound formed by magnesium and sulfur (S) atoms.
ahrayia [7]

Answer:

SIMPLY

( MgS)

6 0
3 years ago
If the percent yield for the following reaction is 75.0%, and 25.0 g of NO₂ are consumed in the reaction, how many grams of nitr
victus00 [196]

Answer:

17.1195 grams of nitric acid are produced.

Explanation:

3NO_2+H_2O\rightarrow 2HNO_3+NO

Moles of nitrogen dioxide :

\frac{25.0 g}{56 g/mol}=0.5434 mol

According to reaction 3 moles of nitrogen dioxides gives 2 moles of nitric acid.

Then 0.5434 moles of nitrogen dioxides will give:

\frac{2}{3}\times 0.5434 mol=0.3623 mol of nitric acid.

Mass of 0.3623 moles of nitric acid :

0.3623 mol\times 63 g/mol=22.8260 g

Theoretical yield = 22.8260 g

Experimental yield = ?

\%Yield=\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

75\%=\frac{\text{Experimental yield}}{22.8260 g}

Experimental yield of nitric acid = 17.1195 g

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Which of the following has the largest radius k,Na+,Na,k+
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Answer: k

Explanation: a p e x

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