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3 years ago

How much heat is required to vaporize 25g of water at 100*c

Chemistry

1 answer:

ElenaW [278]3 years ago

4 0

Answer:

Heat required = 13,325 calories or 55.75 KJ.

Explanation:

To convert a water to steam at 100 degree celsius to vapor, we have to give latent heat of vaporization to water

Which equals ,

Q = mL,

Where, m is the mass of water present

L = specific latent heat of vaporization

Here , m= 25 gram

L equals to 533 calories (or 2230 Joules)

So, Q = 25×533 = 13,325 Calories

Or , Q = 55,750 Joules = 55.75 KJ

so, Heat required = 13,325 calories or 55.75 KJ.

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As an atom's nucleus gets larger the electric charges repelling the protons get larger. To compensate for this greater electric

IrinaK [193]

i think the greater the electric charge the atom decreases in size

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3 years ago

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. Copper(I) oxide, Cu2O, is reduced to metallic copper by heating in a stream of hydrogen gas. What mass of water is produced wh

maksim [4K]

Answer: 0.07868 mol H₂O

Explanation:

1) Chemical equation:

Cu₂O +H₂ → 2Cu + H₂O

2) mole ratios:

1 mol Cu₂O : 1 mol H₂ : 2 mol Cu : 1 mol H₂O

3) Convert 10.00 g of Cu to grams, using the atomic mass:

Atomic mass of Cu: 63.546 g/mol

number of moles = mass in grams / atomic mass = 10.00g / 63.546 g/mol

number of moles = 0.1574 mol

4) Use proportions

2mol Cu 0.1574 mol Cu
--------------- = ---------------------
1 mol H₂O x

⇒ x = 0.1574 mol Cu × 1 mol H₂O / 2mol Cu = 0.07868 mol H₂O

That is the answer

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3 years ago

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Using the Brønsted-Lowry concept of acids and bases, identify the Brønsted-Lowry acid and base in each of the following reaction

Luden [163]

Answer:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

There are no Brønsted-Lowry acids and bases in this reaction.

Explanation:

According to the Brønsted-Lowry concept, when an acid (HA) and a base (B) undergoes a chemical reaction, the acid (HA) loses a proton and forms its conjugate base (A⁻), whereas the base gains (B) the proton to form its conjugate acid (HB⁺).

The chemical equation for this reaction is:

HA + B ⇌ A⁻ + HB⁺

Given reactions:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻

Reason: In this reaction, the acid H₂O loses a proton and forms its conjugate base, OH⁻. Whereas, the base HSO³⁻ gains a proton to form its conjugate acid, H₂SO₃.

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

There are no Brønsted-Lowry acids and bases in this reaction.

Reason: In this reaction, there is no exchange of proton between the acid and the base.

4 0

3 years ago

How mamy moles of NaCl will be produced from 83.0g of Na, assuming Cl2 is available in excess

Anestetic [448]

Answer:

3.62moles

Explanation:

First let us generate a balanced equation for the reaction

2Na + Cl2 —> 2NaCl

Molar Mass of Na = 23g/mol

Mass of Na from the balanced equation = 2 x 23 = 46g

Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol

Mass of NaCl from the balanced equation = 2 x 58.5 = 117g

From the question,

46g of Na produced 117g of NaCl.

Therefore, 83g of Na will produce = (83 x 117)/46 = 211.11g of NaCl

Converting this mass (211.11g of NaCl) to mole, we obtain:

n = Mass /Molar Mass

n = 211.11/ 58.5

3.62moles

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3 years ago

What mass of barium sulfate (233 g/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 125 mL of

Rzqust [24]

Answer:

4.37 g of barium sulphate

Explanation:

The reaction equation is;

3BaCl2(aq) + Fe2(SO4)3(aq) ---->3 BaSO4(s) + 2FeCl3(aq)

From the question, the number of moles of both barium chloride and FeSO4 = 125/1000 L × 0.150 M = 0.01875 moles

To find the limiting reactant;

3 moles of barium chloride yields 3 moles of barium sulphate

0.01875 moles of barium chloride yields 3 × 0.01875 moles/3 = 0.01875 moles of barium sulphate

1 mole of iron III sulphate yields 3 moles of barium sulphate

0.01875 molesof iron III sulphate yields 0.01875 moles ×3/1 = 0.05625 moles of barium sulphate

Hence,barium chloride is the limiting reactant

Amount of barium sulphate produced = 0.01875 moles × 233 g/mol = 4.37 g of barium sulphate

4 0

2 years ago