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3 years ago

Help plz I will give u BRAINLIST

Chemistry

1 answer:

Agata [3.3K]3 years ago

6 0

CS2 + 3O2 = CO2 + 2SO2

1 mole of CS2 gives 1 mole of CO2

12 + 2(32) = 76g of CS2 yields 44 g of CO2

Theoretically 1 g of CS2 yields 44/76 g CO2

Therefore 50 g CS2 should yield 50*44 / 76 = 28.95 g

So % yield = 103.6 % ( which is not possible because you can't create matter from nothing).

The 30g cannot be right . This is experimental err.

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What mass of aluminum is needed to produce 0.500 mole of aluminum chloride?

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Answer: " 13.5 g Al " ;

→ that is: "13.5 grams of aluminum."

____________________________

Explanation:

____________________________

Note: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:

The reactants: "aluminum (Al) " ; and "chlorine (Cl) " ; and:

The product: "aluminum choloride (AlCl₃) " .

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The "balanced chemical equation" is:

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2 Al + 3 Cl₂ → 2 AlCl₃ ;

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Note: The molecular weight of "aluminum (Al)" is: " 26.98 g /mol " .

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So: We call solve using a technique known as: "dimensional analysis" :

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0.500 mol AlCl₃ * $(\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?$

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Note: The units of "mol AlCl₃" cancel out to "1' ; and:

The units of "mol Al" cancel out to "1" ; and we are left with:

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" $\frac{(0.500 * 2 * 26.98)}{2}$ g Al ["grams of aluminum"] ;

____________________________

Note: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:

→ (0.500 * 26.98) g Al ;

= 13.49 g Al ;

→ Round to 3 (Three) significant figures;

→ Since: "0.500" has 3 (Three) significant figures:

____________________________

= 13.5 g Al ; that is: "13.5 grams of aluminum."

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Hope this is helpful!

Best wishes to you in your academic pursuits—and within the "Brainly" community!

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3 0

3 years ago

A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo

Liula [17]

Answer: The molar mass of the insulin is 6087.2 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol$

Hence, the molar mass of the insulin is 6087.2 g/mol

8 0

3 years ago