2 years ago

What volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B)?

Chemistry

1 answer:

vova2212 [387]2 years ago

6 0

Idk what is is i am sorry

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A beaker is filled to the 500 mL mark with alcohol. What increase in volume (in mL) does the beaker contain when the temperature

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Answer:

"1.4 mL" is the appropriate solution.

Explanation:

According to the question,

$v_0=500$
$\alpha =1.12\times 10^{-4}$
$\Delta \epsilon = 25$

Now,

Increase in volume will be:

⇒ $\Delta V = \alpha\times v_0\times \Delta \epsilon$

By putting the given values, we get

$=1.12\times 10^{-4}\times 500\times 25$

$=1.12\times 10^{-4}\times 12500$

$=1.4 \ mL$

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Answer:

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How much energy (kJ) is required to change 0.18 mole of ice (s) at 0 C to water (l) at 0 C?

Dmitriy789 [7]

Answer:25,06 kJ of energy must be added to a 75 g block of ice.

ΔHfusion(H₂O) = 6,01 kJ/mol.

T(H₂O) = 0°C.

m(H₂O) = 75 g.

n(H₂O) = m(H₂O) ÷ M(H₂O).

n(H₂O) = 75 g ÷ 18 g/mol.

n(H₂O) = 4,17 mol.

Q = ΔHfusion(H₂O) · n(H₂O)

Q = 6,01 kJ/mol · 4,17 mol

Q = 25,06 kJ.

Explanation:

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