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3 years ago

Given a 1.50 Liters of solution that contains 62.5 grams of magnesium oxide. What is the Molarity of that solution?

1 answer:

5 0

RMM of magnesium oxide=24+16=40
40g=1mol
therefore 62.5g will have =62.5/40=1.5625 moles
1.5 litres contain 1.5625 mols
therefore 1 litre =1*1.5625/1.5=1.042M

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What mass in grams would 5.7L of hydrogen gas occupy at STP?

tekilochka [14]

Answer: The correct answer is: " 0.54 g " .

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Explanation:

Note that "hydrogen gas" is:

H₂ (g) ; that is: a "diatomic element" (diatomic gas) ;

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The molecular weight of "H" is: 1.00794 g ;

(From the Periodic Table of Elements).

So, the molecular weight of: H₂ (g) is:

" 1.00794 g * 2 = 2.01588 g ; {use calculator) ;

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Note the conversion for a gas at STP:

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1 mol of a gas = 22.4 L gas;

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i.e. " 1 mol / 22.4 L " ;

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So: " 5.7 L H₂ (g) $* \frac{1 mol H_{2} }{22.4 L} *\frac{2.01588 g}{mol} =? ;$

The "L" ("literes" cancel out to "1" ; since "L/L = 1 ;

The "mol" (moles) cancel out to "1" ; since "mol/mol = 1 ;

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and we are left with:

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[5.7 * 2.104588 g ] / 22.4 = ? g ;

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→ [ 11.9961516 g ] / 22.4 =

0.53554248214 g ;l

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We round this value to: " 0.54 g " ;

→ since "5.7 L " has 2 (two) significant figures;

22.4 is an exact number conversion;

and "5.7 L" has fewer significant figures than:

" 2.104588 " ; or: " 1.00794 " .

→ as such: We round to "2 (two) significant figures."

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Hope this is helpful. Wishing you the best in your academic endeavors!

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3 years ago

Write the names of stages in the life cycle of butterfly

stepladder [879]

Answer:

There are four stages in the metamorphosis of butterflies and moths: egg, larva, pupa, and adult.

Hope this helps :)

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3 years ago

An experiment shows that a 236 mL gas sample has a mass of 0.443 g at a pressure of 740 mmHg and a temperature of 22 ∘C. What is

aleksley [76]

Answer:

49.2 g/mol

Explanation:

Let's first take account of what we have and convert them into the correct units.

Volume= 236 mL x ( $\frac{1 L}{1000 mL}$ ) = .236 L

Pressure= 740 mm Hg x ( $\frac{1 atm}{760 mm Hg}$ )= 0.97 atm

Temperature= 22C + 273= 295 K

mass= 0.443 g

Molar mass is in grams per mole, or MM= $\frac{mass}{moles}$ or MM= $\frac{m}{n}$ . They're all the same.

We have mass (0.443 g) we just need moles. We can find moles with the ideal gas constant PV=nRT. We want to solve for n, so we'll rearrange it to be

n= $\frac{PV}{RT}$ , where R (constant)= 0.082 L atm mol-1 K-1

Let's plug in what we know.

n= $\frac{(0.97 atm)(0.236 L)}{(0.082)(295K)}$

n= 0.009 mol

Let's look back at MM= $\frac{m}{n}$ and plug in what we know.

MM= $\frac{0.443 g}{0.009 mol}$

MM= 49.2 g/mol

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Explanation:

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Answer:

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