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3 years ago

How much energy is required to melt 76.941g of ice

Chemistry

1 answer:

olga55 [171]3 years ago

5 0

Answer:

Q=25.7 Kj

Explanation:

76.941 g H2O*1 mol/18.016= 4.27 Mol H20

(4.27 Mol H2O)(6.009 Kj/Mol)

Q=25.7 Kj

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How many O2 molecules react with 47 CH4 molecules according to the preceding equation CH4 + O2 → CO2 + H2O

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3 years ago

At the equivalence point of a titration of the [H+] concentration is equal to:

icang [17]

B. At the equivalence point of a titration of the [H+] concentration is equal to 7.

<h3>What is equivalence point of a titration?</h3>

The equivalence point of a titration is a point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution.

At the equivalence point in an acid-base titration, moles of base equals moles of acid and the solution only contains salt and water.

At the equivalence point, equal amounts of H+ and OH- ions combines as shown below;

H⁺ + OH⁻ → H₂O

The pH of resulting solution is 7.0 (neutral).

Thus, the pH at the equivalence point for this titration will always be 7.0.

Learn more about equivalence point here: brainly.com/question/23502649

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1 year ago

What inference can be drawn from the graph?

defon

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1 year 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

3 years ago