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3 years ago

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Copper has a specific heat of 0,900 J(g C). How much energy in kJ is needed to raise the temperature of a 700 g block of aluminu

m from 30.7°C to 82,1C?(show your steps)

1 answer:

sashaice [31]3 years ago

4 0

Answer:

32.4 kJ

Explanation:

Given that:

Specific heat capacity fo copper, c = 0.9

Mass of copper, m = 700 g

Temperature change, dθ ;t2 - t1 = (82.1 - 30.7) = 51.4

Energy required, E = mcdθ

E = 700 * 0.9 * 51.4

E = 32382 J

Energy required on kJ

32382 / 1000

= 32.382 kJ

= 32.4 kJ

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How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)

Nutka1998 [239]

Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = $\frac{100 g}{44 g/mol}=2.273 moles$

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = $2.273\times 10^{-3} kmol$

2) 1 liter of ethyl alcohol of density $0.789 g/cm^3$

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = $0.789 g/cm^3$

$1 cm^3=1 mL$

Mass of ethyl alcohol = m

$m=d\times V=0.789 g/cm^3\times 1000 mL=789 g$

Molar mass of ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = $\frac{789 g}{46 g/mol}=17.152 mol$

$17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol$

3) Volume of oxygen gas,V = $1.5 m^3=1500 L$

$1 m^3= 1000 L$

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

$PV=nRT$

$n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol$

0.01632 mol = 0.01632 × 0.001 kmol= $1.632\times 10^{-5} kmol$

4) Volume water in mixture = 1 L

Density of water = $1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L$

Mass of water = $1000 g/L\times 1 L = 1000 g$

Volume of alcohol = 2.5 L

Density of alcohol = $789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L$

Mass of alcohol = $789 g/L\times 2.5 L = 1972.5 g$

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

$\frac{1000 g}{2972.5 g}\times 100=33.64\%$

Mass percentage of alcohol :

$\frac{1972.5 g}{2972.5 g}\times 100=66.36\%$

Moles of water :

$n_1=\frac{1000 g}{18 g/mol}=55.55 mol$

Moles of alcohol =

$n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol$

Mole fraction of water :

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644$

Mole fraction of alcohol :

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356$

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Find the number of moles of the following substances 210 grams of NaHCO3

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The number of moles of 210 grams of NaHCO₃ is 2.5 moles.

<h3>How to find the Number of moles ? </h3>

To calculate the number of moles use the formula

Number of moles = $\frac{\text{Given Mass}}{\text{Molar Mass}}$

Mass of NaHCO₃ = 210 g

Now we have to find the Molar mass of NaHCO₃

= Atomic mass of Na + Atomic mass of H + Atomic mass of C + 3 (Atomic mass of O)

= 23 + 1 + 12 + 3 (16)

= 36 + 48

= 84 g/mol

Now put the value in above formula we get

Number of moles = $\frac{\text{Given Mass}}{\text{Molar Mass}}$

= $\frac{210\ g}{84\ g/mol}$

= 2.5 moles

Thus from the above conclusion we can say that The number of moles of 210 grams of NaHCO₃ is 2.5 moles.

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