Please help!!

how much heat is required to raise the temperature of a 5.45-g sample of iron (specific heat = 0.450 j/g°c) from 25.0°c to 79.8°

Mamont248 [21]

Answer:

134.397 Joules

Explanation:

Using the formula:

E = C × m × Δθ (where E is Energy, C is specific heat capacity and Δθ is change in temperature)

So E = 0.45×5.45×(79.8-25)

So E = 134.397 Joules

5 0

3 years ago

A buffer solution contains 0.240 M ammonium chloride and 0.499 M ammonia. If 0.0565 moles of perchloric acid are added to 250 mL

snow_lady [41]

Answer:

The pH of the resulting solution is 9.02.

Explanation:

The initial pH of the buffer solution can be found using the Henderson-Hasselbalch equation:

$pOH = pKb + log(\frac{[NH_{4}Cl]}{[NH_{3}]})$

$pOH = -log(1.78\cdot 10^{-5}) + log(\frac{0.240}{0.499}) = 4.43$

$pH = 14 - pOH = 14 - 4.43 = 9.57$

Now, the perchloric acid added will react with ammonia:

$n_{NH_{3}} = 0.499moles/L*0.250 L - 0.0565 moles = 0.0683 moles$

Also, the moles of ammonium chloride will increase in the same quantity according to the following reaction:

NH₃ + H₃O⁺ ⇄ NH₄⁺ + H₂O

$n_{NH_{4}Cl} = 0.240 moles/L*0.250L + 0.0565 moles = 0.1165 moles$

Finally, we can calculate the pH of the resulting solution:

$pOH = pKb + log(\frac{[NH_{4}Cl]}{[NH_{3}]})$

$pOH = -log(1.78\cdot 10^{-5}) + log(\frac{0.1165 moles/0.250 L}{0.0683 moles/0.250 L}) = 4.98$

$pH = 14 - 4.98 = 9.02$

Therefore, the pH of the resulting solution is 9.02.

I hope it helps you!

5 0

3 years ago

Which is the empirical formula for hydrogen peroxide

KiRa [710]

H2O2 i think this is it

8 0

4 years ago

Read 2 more answers

2.5 moles of sodium chloride is dissolved to make 0.050 liters of solution

Hitman42 [59]

The answer is:

the molarity = 50 moles/liters

The explanation:

when the molarity is = the number of moles / volume per liters.

and when the number of moles =2.5 moles

and the volume per liters = 0.05 L

so by substitution:

the molarity = 2.5moles/0.05L

= 50 moles /L

7 0

3 years ago

Calculate the volume of carbon dioxide at 20.0°C and 0.941 atm produced from the complete combustion of 4.00 kg of methane. Comp

tankabanditka [31]

Answer:

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6390.89 Liters.

More volume of carbon dioxide gas was released on combustion of 4.00 kg of propane in comparison to the volume of carbon dioxide released on combustion of 4.00 kg of methane.

Explanation:

Methane

$CH_4+2O_2\rightarrow CO_2+2H_2O$

Mass of methane = 4.00 kg = 4000 g (1 kg = 1000 g)

Moles of methane = $\frac{4000 g}{16 g/mol}=250 mol$

According to reaction, 1 mole of methane gives 1 mole of carbon dioxide gas,then 250 moles of methane will give :

$\frac{1}{1}\times 250 mol=250 mol$ of carbon dioxide gas

Moles of carbon dioxide gas = n = 250 mol

Pressure of carbon dioxide gas = P = 0.941 atm

Temperature of carbon dioxide gas = T = 20.0°C = 20.0+273 K = 293 K

Volume of carbon dioxide gas = V

$PV=nRT$ (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{250 mol\times 0.0821 atm L/mol K\times 293 K}{0.941 atm}=6390.89 L$

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6390.89 Liters.

Propane

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Mass of propane = 4.00 kg = 4000 g (1 kg = 1000 g)

Moles of propane = $\frac{4000 g}{44 g/mol}=90.91 mol$

According to reaction, 1 mole of propane gives 3 mole of carbon dioxide gas,then 90.91 moles of propane will give :

$\frac{3}{1}\times 90.91 mol=272.73 mol$ of carbon dioxide gas

Moles of carbon dioxide gas = n = 272.73 mol

Pressure of carbon dioxide gas = P = 0.941 atm

Temperature of carbon dioxide gas = T = 20.0°C = 20.0+273 K = 293 K

Volume of carbon dioxide gas = V

$PV=nRT$ (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{272.73 mol\times 0.0821 atm L/mol K\times 293 K}{0.941 atm}=6,971.95 L$

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6,971.95 Liters.

Volume of carbon dioxide given by combustion of 4.00 kg of methane = 6390.89 Liters.

Volume of carbon dioxide given by combustion of 4.00 kg of propane = 6,971.95 Liters.

6390.89 Liters < 6,971.95 Liters

More volume of carbon dioxide gas was released on combustion of 4.00 kg of propane in comparison to the volume of carbon dioxide released on combustion of 4.00 kg of methane.

7 0

3 years ago