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3 years ago

The enthalpy of vaporization of water is 2,257,000 J/kg. If I have a 1 kg sample, how much energy is needed to boil all of it

Chemistry

1 answer:

worty [1.4K]3 years ago

8 0

Answer:

2257000 J

Explanation:

Applying,

Q = Cₓm.................. Equation 1

Where Q = amount of energy need to boil the water, Cₓ = Enthalpy of vaporization of water, m = mass of water.

From the question,

Given: Cₓ = 2257000 J/kg, m = 1 kg

Substitute these values into equation 1

Q = 2257000×1

Q = 2257000 J

Hence the energy needed to boil all of the water is 2257000 J

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What term expresses the ratio of moles of reactants and products according to the coefficients in the balanced chemical equation

bogdanovich [222]

Answer:

A) mole ratio

Explanation:

<em>A mole ratio</em>, also known as a mole-to-mole ratio, <em>is the ratio between the amounts in moles of one reactant/product to the moles of the other reactant/product. </em>This ratio is determined considering the coefficients in a balanced chemical equation. This ratio is used in chemical problems as a conversion factor between the compounds involved in the reaction.

I hope you find this information useful and interesting! Good luck!

6 0

3 years ago

Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the fi

Savatey [412]

The question is incomplete, the complete question is;

Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the first heating, your crucible and contents weighs 17.51 g. After the second heating, your crucible and contents weighs 17.50 g.

What is the theoretical yield of sodium carbonate?

What is the experimental yield of sodium carbonate?

What is the percent yield for sodium carbonate?

Which errors could cause your percent yield to be falsely high, or even over 100%?

Answer:

See Explanation

Explanation:

We have to note that water is driven away after the second heating hence we are concerned with the weight of the pure dry product.

Hence;

From the reaction;

2 NaHCO3 → Na2CO3(s) + H2O(l) + CO2(g)

Number of moles of sodium bicarbonate = 18.56 - 15.98 = 2.58 g/87 g/mol

= 0.0297 moles

2 moles of sodium bicarbonate yields 1 mole of sodium carbonate

0.0297 moles of 0.015 moles sodium bicarbonate yields 0.0297 * 1/2 = 0.015 moles

Theoretical yield of sodium carbonate = 0.015 moles * 106 g/mol = 1.59 g

Experimental yield of sodium bicarbonate = 17.50 g - 15.98 g = 1.52 g

% yield = experimental yield/Theoretical yield * 100

% yield = 1.52/1.59 * 100

% yield = 96%

The percent yield may exceed 100% if the water and CO2 are not removed from the system by heating the solid product to a constant mass.

5 0

2 years ago

A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base

tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ = 0.00375 mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

$HN_3 + OH- ---> N^-_{3} + H_2O$

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L

Concentration of $HN_3$ = $\dfrac{0.001755}{0.0383}$ = 0.0458 M

Concentration of $N^{-}_3$ = $\dfrac{ 0.001995 }{0.0383}$ = 0.0521 M

GIven that :

Ka = $1.9 x 10^{-5}$

Thus; it's pKa = 4.72

$pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})$

$pH =4.72 + log(1.1376)$

$pH =4.72 + 0.05598$

$pH =4.77598$

pH ≅ 4.80

3 0

3 years ago

Which of these elements has the smallest atomic radius?

nadya68 [22]

Answer:

3. be

Explanation:

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3 years ago

Anyone have a good tip, or app to help me with nomenclature???

ExtremeBDS [4]

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