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3 years ago

Which equation is used to help form the combined gas law?

Chemistry

1 answer:

padilas [110]3 years ago

6 0

<h3>Answer:</h3>

The equation is; P₁V₁/T₁ = P₂V₂/T₂

<h3>Explanation:</h3>

Concept being tested: Combined gas law

We need to know the two gas laws that form the combined gas law

The combined gas laws is derived from the Boyle's gas law and Charles's law.

According to Boyle's law;

P₁V₁ = P₂V₂

On the other hard, according to Charles's law;

V₁/T₁= V₂/T₂

The combine gas laws combines both Boyle's law and Charles's law.

According to the combined gas law, the volume of a fixed mass of a gas i inversely proportional to pressure and directly proportional to its absolute temperature.

Therefore, combining both Boyle's law and Charles's law, we get;

P₁V₁/T₁ = P₂V₂/T₂

Thus, based on Combined gas law, then

P₁V₁/T₁ = P₂V₂/T₂

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Calculate the moles and grams of solute in each solution. D. 2.0 L of 0.30M Na2SO4. I already have A, B, and C. Thanks!

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Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
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n = C * V
n = 0.30 mol / L * 2 L
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A sample of nitrogen gas has the temperature drop from 250.°c to 150.°c at constant pressure. what is the final volume if the in

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It states that for a fixed amount of gas, temperature is directly proportional to volume of gas.
V / T = k
where V- volume , T - temperature and k - constant
$\frac{V1}{T1} = \frac{V2}{T2}$
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation.
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4 years ago

Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small a

GrogVix [38]

Answer:

The calorie content of 65g of candy is 326.78 cal

Explanation:

Step 1: Data given

Mass of the candy = 15.00 grams

Mass of the container = 0.325 kg

Mass of water = 1.75kg

0.624 kg at an initial temperature of 15.0°C.

The specific heat of aluminium = 0.22 Cal/kg°C

The specific heat of water = 1 cal/kg°C

Step 2: Calculate calorie content for a 15 gram sample

ΔQ = Σm*c*ΔT

⇒ m = mass in grams

⇒ with c= the specific heat in Cal/kg°C

⇒ with ΔT = T2 -T1 = the change in temperatures in °C

ΔQ = m(bomb) * C(aluminium) * ΔT + m(cup) * C(aluminium) * ΔT + m(H2O) * c(H20) * ΔT

ΔQ = (m(bomb) + m(cup)) * c(aluminium) + m(H2O)*c(H20) ) * ΔT

⇒ with mass of the bomb calorimeter = 0.325 kg

⇒ with mass of the cup = 0.624 kg

⇒ with c(aluminium) = the specific heat of aluminium = 0.22 Cal/kg°C

⇒ with mass of water = 1.75 kg

⇒ with c(water) = the heat capacity of water = 1 Cal/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 53.5 - 15.0 = 38.5 °C

ΔQ = 0.325*0.22*38.5 + 0.624*0.22*38.5 + 1.75*1*38.5

ΔQ = ((0.325 + 0.624)*0.22 + 1.75*1)*38.5

ΔQ = 75.41 cal

Step 3: Calculate the calorie content for a 65 gram sample

For a 65g sample the calorie content will be more or less 4x higher than a 15 gram sample:

ΔQ = 75.41 * (65/15) = 326.78 cal

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