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xz_007 [3.2K]
3 years ago
11

If the vapor's volume were to be incorrectly recorded as 125ml, how will this error affect the calculated molar mass of the unkn

own?
Chemistry
1 answer:
pishuonlain [190]3 years ago
3 0
Since you didn't give the actual volume (or any of the experimental values) I can only tell you how to do it. Do the calculation using the real (determined) volume of the flask. Then, re-do the calculation with v = 125ml. Take the two values and calculate % error; m = measured vol; g = guessed vol. 

<span>[mW (m) - mW (g)]/mW (m) x 100% </span>

<span>(they want % error so, if it is negative, just get rid of the sign) </span>
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Mg(OH)2 +2HCl---> MgCl2+2H2O
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2 years ago
Reversible reaction and examples on them
Tomtit [17]

Answer:

Here are three examples  

Explanation:

In a reversible reaction, the conversions of reactants to products and of products to reactants occur at the same time.

Example 1

The reaction of hydrogen and iodine to from hydrogen iodide.  

H₂ + I₂ ⇌ 2HI

Example 2

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4 0
3 years ago
Ice, liquid, and water vapor are the three states of water
UNO [17]
This is true. Solid, Liquid, and Gas.


Hope that helps!!
7 0
2 years ago
Select all the statements that correctly describe a solution. Multiple select question.
motikmotik

Answer:

B, D, C

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8 0
2 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]

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

<u>Explanation:</u>

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
2 years ago
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