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2 years ago

How does changing the initial temperature of the copper affect how much heat energy it has?

Chemistry

1 answer:

taurus [48]2 years ago

5 0

Answer:

the rock has a greater amount of heat energy which transfers to water causing vaporization.

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A stock solution is made by dissolving 66.05 g of (NH4)2SO4 in enough water to make 250 mL of solution. A 10.0 mL sample of this

Degger [83]

If the solute is properly distributed in the given volume, there are 2.642 g of (NH4)2SO4 per 10 mL. For the new solution, divide the 2.642 g by the molar mass of the compound. The answer is 0.02 moles. Then, divide this by the new volume, 50 mL or 0.05 L. The concentration of the new solution is 0.4 M.

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3 years ago

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What is the role of critical thinking in scientific process

laiz [17]

Answer:

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3 years ago

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Using the equation, C5H12 + 8O2 Imported Asset 5CO2 + 6H2O, if 2 moles of pentane (C5H12) were supplied, and an unlimited amount

Kamila [148]

12 moles of water H₂O are produced from the combustion of pentane.

Explanation:

We have the following combustion of pentane (C₅H₁₂):

C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

Knowing the chemical reaction we devise the following reasoning:

if 1 moles of pentane C₅H₁₂ produces 6 moles of water H₂O

then 2 moles of pentane C₅H₁₂ produces X moles of water H₂O

X = (2 × 6) / 1 = 12 moles of water H₂O

Learn more about:

combustion of organic compounds

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3 years ago

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

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3 years ago

One method of increasing the chances of particles colliding during a reaction is to...

Pepsi [2]

Answer:

Explanation:

Increasing concentration

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3 years ago

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