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3 years ago

A gas occupies 900.0 ML at temperatures of 27.0 Celsius. What is the volume at 123.0 celsius

Chemistry

2 answers:

Artemon [7]3 years ago

4 0

Answer:

1188.0 mL.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have two different values of V and T:

V₁T₂ = V₂T₁

V₁ = 900 mL, T₁ = 27.0°C + 273 = 300.0 K.

V₂ = ??? mL, T₂ = 123.0°C + 273 = 396.0 K.

∴ V₂ = V₁T₂/T₁ = (900 mL)(396 K)/(300.0 K) = 1188.0 mL.

Tamiku [17]3 years ago

3 0

Answer:

1188 ml

Explanation:

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3 years ago

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3 years ago

How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 26 ∘C to 6 ∘C? (Assume the density o

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Answer:

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

Explanation:

Heat gain by ice = Heat lost by water

Thus,

Heat of fusion + $m_{ice}\times C_{ice}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)$

Where, negative sign signifies heat loss

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Heat of fusion + $m_{ice}\times C_{ice}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)$

Heat of fusion = 334 J/g

Heat of fusion of ice with mass x = 334x J/g

For ice:

Mass = x g

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Final temperature = 6 °C

Specific heat of ice = 1.996 J/g°C

For water:

Volume = 353 mL

$Density (\rho)=\frac{Mass(m)}{Volume(V)}$

Density of water = 1.0 g/mL

So, mass of water = 353 g

Initial temperature = 26 °C

Final temperature = 6 °C

Specific heat of water = 4.186 J/g°C

So,

$334x+x\times 1.996\times (6-0)=353\times 4.186\times (26-6)$

$334x+x\times 11.976=29553.16$

345.976x = 29553.16

x = 85.4197 kg

Thus,

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

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3 years ago