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3 years ago

How many moles of sodium are present in 17 g of Na?

Chemistry

1 answer:

11Alexandr11 [23.1K]3 years ago

8 0

Answer:

1. 0.74mol

2. 0.42mol

3. 2.125mol

4. 0.301mol

5. 4.52 × 10^23 particles

Explanation:

Number of moles (n) in a substance can be found using the formula:

mole (n) = mass/molar mass

Using this formula, the following moles are calculated:

1. Molar of Na = 23g/mol

mole = 17/23

mole = 0.74mol

2. Molar mass of Na2SO4 = 23(2) + 32 + 16(4)

= 46 + 32 + 64

= 142g/mol

Mole = 60/142

mole = 0.42mol

3. Molar mass of CO2 = 12 + 16(2)

= 12 + 32

= 44g/mol

mole = 93.5/44

mole = 2.125mol

4. Molar mass of sodium nitrate (NaNO3) = 23 + 14 + 16(3)

= 23 + 14 + 48

= 85g/mol

mole = 25.6/85

mole = 0.301mol

5. Number of particles in one mole of a substance is 6.022 × 10^23 particles. Hence, in 0.75mol of calcium hydroxide (Ca(OH)2, there will be;

0.75mol × 6.02 × 10^23

= 4.515 × 10^23

= 4.52 × 10^23 particles

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Answer:

Explanation:

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LiCl(aq) + AgNO₃(aq) → AgCl(s) + LiNO₃(aq)

Ionic equation:

Li⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) + Li⁻(aq) + NO⁻₃(aq)

Net ionic equation:

Cl⁻(aq) + Ag⁺(aq) → AgCl(s)

C = H2SO4(aq)+Li2SO3(aq)→

Chemical equation:

H₂SO₄(aq) + Li₂SO₃(aq) → Li₂SO₄(aq) + SO₂(g) + H₂O(l)

Ionic equation:

2H⁺(aq) + SO²⁻₄(aq) + 2Li⁺(aq) + SO₃²⁻(aq) → 2Li⁺ (aq) + SO₄²⁻(aq) + SO₂(g) + H₂O(l)

Net ionic equation:

2H⁺ + SO₃²⁻(aq) → SO₂(g) + H₂O(l)

Part E: HClO4(aq)+Ca(OH)2(aq)→

Chemical equation:

HClO₄(aq) + Ca(OH)₂(aq) → Ca(ClO₄)₂ (aq) + H₂O(l)

Balanced Chemical equation:

2HClO₄(aq) + Ca(OH)₂(aq) → Ca(ClO₄)₂ (aq) + 2H₂O(l)

Ionic equation:

2H⁺(aq) + 2ClO⁻₄(aq) + Ca²⁺(aq) + (OH)²⁻₂(aq) → Ca²⁺(aq) +(ClO₄)²⁻₂ (aq) + 2H₂O(l)

Net ionic equation:

2H⁺(aq) + (OH)²⁻₂(aq) → 2H₂O(l)

Part F: Cr(NO3)3(aq)+LiOH(aq)→

Chemical equation:

Cr(NO₃)₃(aq) + LiOH (aq) → LiNO₃(aq) + Cr(OH)₃(s)

Balanced chemical equation;

Cr(NO₃)₃(aq) + 3LiOH (aq) → 3LiNO₃(aq) + Cr(OH)₃(s)

Ionic equation:

Cr³⁺(aq) + 3NO₃⁻(aq) + 3Li⁺(aq) + 3OH⁻ (aq) → 3Li⁺(aq) + 3NO⁻₃(aq) + Cr(OH)₃(s)

Net ionic equation:

Cr³⁺(aq) + 3OH⁻ (aq) → Cr(OH)₃(s)

Part H: HCl(aq)+Hg2(NO3)2(aq)→

Chemical equation:

HCl (aq) + Hg₂(NO₃)₂(aq) → Hg₂Cl₂ (s) + HNO₃(aq)

Balanced chemical equation:

2HCl (aq) + Hg₂(NO₃)₂(aq) → Hg₂Cl₂ (s) + 2HNO₃(aq)

Ionic equation;

2H⁺(aq) + 2Cl⁻ (aq) + 2Hg⁺(aq) + 2NO₃⁻(aq) → Hg₂Cl₂ (s) + 2H⁺(aq) + 2NO⁻₃(aq)

Net ionic equation:

2Cl⁻ (aq) + 2Hg⁺(aq) → Hg₂Cl₂ (s)

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3 years ago

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C. Neither a Physical or Chemical Changu

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I think B a chemical change

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A methanol-water mixture is to be flash distilled at 1 atm. If the feed is 25 mole %methanol, what are the liquid and vapor comp

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Answer:

Explanation:

Given that:

The distillation is carried out at a pressure of 1 atm

The feed harbors 25% mole of methanol (z)

The total moles of feed is usually 100 moles

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Using the total mole balance for the distillation column.

Fz = Lx + Vy

where;

F = amount of feed

z = mole fraction of ethanol (in feed)

L = amount of liquid product out of the column

V = amount of vapor product out of the column

x = mole fraction of methanol out of the liquid

y = mole fraction of methanol out of the vapor.

SO;

(a)

If all the feed is vaporized, then the vapor will likely have the same composition as the feed.

(b)

If no vaporization of the feed takes place, then the bottoms moving out of the column contains the same composition as the feed.

(c)

If 1/3 of the feed is vaporized; then 2/3 of the feed is at the bottom.

The balance equation would be:

$Fz = (\dfrac{2}{3}F) x + (\dfrac{1}{3}F)y \\ \\ z = \dfrac{2}{3}x+\dfrac{1}{3}y$

Replacing z = 0.25; we have:

$0.25 = \dfrac{2}{3}x+\dfrac{1}{3}y$

0.75 = 2x + y

(d)

If 2/3 of the feed is vaporized;

Then:

$Fz = (\dfrac{1}{3}F) x + (\dfrac{2}{3}F)y \\ \\ z = \dfrac{1}{3}x+\dfrac{2}{3}y$

replacing z = 0.25

$0.25 = \dfrac{1}{3}x+\dfrac{2}{3}y$

0.75 = x + 2y

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Answer:

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Explanation:

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