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Butoxors [25]
4 years ago
13

What is the molality of a solution in which 0.32 moles AlCl3 has been dissolved in 2,200 g water? What mass of water is needed t

o prepare a 1.20 molal solution using 0.60 mol propyleneglycol? What is the molality of a solution in which 0.145 mol CO2 is dissolved in 591 g water?
Chemistry
1 answer:
klemol [59]4 years ago
8 0

Answer:

1) The molality is 0.145 molal

2) We need 0.50 kg water

3) The molality is 0.245 molal

Explanation:

What is the molality of a solution in which 0.32 moles AlCl3 has been dissolved in 2,200 g water?

Molality = moles AlCl3 / mass water (in kg)

Molality = 0.32 moles / 2.2 kg water

Molality = 0.145 moles/kg = 0.145 molal

The molality is 0.145 molal

What mass of water is needed to prepare a 1.20 molal solution using 0.60 mol propyleneglycol?

Molality = moles propyleneglycol / mass water

1.20 = 0.60 / x kg

X = 0.60 / 1.2

X = 0.50 kg water

We need 0.50 kg water

What is the molality of a solution in which 0.145 mol CO2 is dissolved in 591 g water?

Molality = moles CO2 / mass water (in kg)

Molality = 0.145 moles / 0.591 kg

Molality = 0.245 molal

The molality is 0.245 molal

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How many grams of ethylene glycol (c2h6o2 must be added to 1.25 kg of water to produce a solution that freezes at -5.88 ∘c?
e-lub [12.9K]
The freezing point depression is calculated through the equation,
                                    ΔT = (kf)  x m 
where ΔT is the difference in temperature, kf is the freezing point depression constant (1.86°C/m), and m is the molality. Substituting the known values,
                                   5.88 = (1.86)(m)
m is equal to 3.16m

Recall that molality is calculated through the equation,
                                  molality = number of mols / kg of solvent
                                       number of mols = (3.16)(1.25) = 3.95 moles
Then, we multiply the calculated amount in moles with the molar mass of ethylene glycol and the answer would be 244.9 g.

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3 years ago
PLZ HELP, BRAINLIEST TO WHOEVER GIVES WORK TOO!!​
Flura [38]

Answer:

1.15 hours

Explanation:

If they are going 100 kilometers an hour, and they need to go 115 k, then do 115/100 which is 1.15 which is the time

5 0
3 years ago
How much time does it take for light from the center of the galaxy to reach Earth?
Ede4ka [16]

Answer:

Light moves at 300,000 kilometers per second, divide these and you get 500 seconds, or 8 minutes and 20 seconds this is an average number.

Explanation:

3 0
3 years ago
Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magne
Fynjy0 [20]

Fireworks owe their colors to reactions of combustion of the metals present. When Mg and Al burn, they emit a white bright light, whereas iron emits a gold light. Besides metals, oxygen is necesary for the combustion. The decomposition reactions of barium nitrate and potassium chlorate provide this element. At the same time, barium can burn emitting a green light.

(a) Barium nitrate is a <em>salt</em> formed by the <em>cation</em> barium Ba²⁺ and the <em>anion</em> nitrate NO₃⁻. Its formula is Ba(NO₃)₂. Potassium chlorate is a <em>salt</em> formed by the <em>cation</em> potassium K⁺ and the <em>anion</em> chlorate ClO₃⁻. Its formula is KClO₃.

(b) The balanced equation for the decomposition of potassium chloride is:

2KClO₃(s) ⇄ 2KCl(s) + 3O₂(g)

(c)  The balanced equation for the decomposition of barium nitrate is:

Ba(NO₃)₂(s) ⇄ BaO(s) + N₂(g) + 3O₂(g)

(d) The balanced equations of metals with oxygen to form metal oxides are:

  • 2 Mg(s) + O₂(g) ⇄ 2 MgO(s)
  • 4 Al(s) + 3 O₂(g) ⇄ 2 Al₂O₃(s)
  • 4 Fe(s) + 3 O₂(g) ⇄ 2 Fe₂O₃(s)

5 0
4 years ago
In one sample of a compound of copper and oxygen, 3.12g of the compound contains 2.50g of copper and the remainder is oxygen. Ca
blsea [12.9K]

Answer:

% composition O = 19.9%

% composition Cu = 80.1%

Explanation:

Given data:

Total mass of compound = 3.12 g

Mass of copper = 2.50 g

Mass of oxygen = 3.12 - 2.50 = 0.62 g

% composition = ?

Solution:

Formula:

<em>% composition = ( mass of element/ total mass)×100</em>

% composition Cu = (2.50 g / 3.12 g)×100

% composition Cu = 0.80 ×100

% composition Cu = 80.1%

For oxygen:

<em>% composition = ( mass of element/ total mass)×100</em>

% composition O = (0.62 g / 3.12 g)×100

% composition O = 0.199 ×100

% composition O = 19.9%

5 0
3 years ago
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