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emmainna [20.7K]
3 years ago
13

385 J of heat are needed to heat a piece of aluminum from 22°C to 45°C. If the specific heat of Al is 0.90 J/g°C, what is the ma

ss of the aluminum sample?
Pls show work :))
Chemistry
1 answer:
VikaD [51]3 years ago
4 0

Answer:

M of Al=33.09g or 0.0331kg

Explanation:

Heat Energy= specific heat*mass*change in temperature

H=M*C*T

make M subject of the formula

M=H/CT

M=685J/0.90J/g°C*(45°C-22°C)

M=685J/0.90J/g°C*23°C

M=685J/20.7J/g

M=33.09g or 0.0331kg

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Calculate the solubility of carbon dioxide at 400 kPa.
BaLLatris [955]

The solubility of carbon dioxide at 400 kPa  at room temperature is ;

( B ) 0.61 CO2/L

<u>Given data </u>

pressure of CO₂ = 400 Kpa = 3.95 atm

Kh of CO₂ = 3.3 * 10⁻² mol/L.atm

<h3>Calculate the solubility of carbon dioxide </h3>

Solubility = pressure * Kh value of CO₂

                = 3.95 atm * 3.3 * 10⁻² mol / L.atm

                = 0.13 mol/l  CO₂

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Hence we can conclude that the solubility of CO₂ at 400 kPa is 0.13 mol/l  CO₂.

Learn more about solubility : brainly.com/question/23946616

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2 years ago
Ana drives at a speed of 175m/s. she recorded a distance of 2km. Find the time.
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To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w
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<u>Answer:</u> The number of moles of weak acid is 4.24\times 10^{-3} moles.

<u>Explanation:</u>

To calculate the moles of KOH, we use the equation:

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We are given:

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Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol

The chemical reaction of weak monoprotic acid and KOH follows the equation:

HA+KOH\rightarrow KA+H_2O

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, 4.24\times 10^{-3}mol of KOH will react with = \frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol of weak monoprotic acid.

Hence, the number of moles of weak acid is 4.24\times 10^{-3} moles.

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