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Iteru [2.4K]
2 years ago
7

When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier t

han ordinary seawater) is evaporated, 151 grams of MgCl2 are recovered. What is the molarity of MgCl2 in the original sample?
Chemistry
1 answer:
lisov135 [29]2 years ago
8 0

Answer:

1.59mol/L

Explanation:

Data obtained from the question include:

Mass of MgCl2 = 151g

Volume of water(solvent) = 1L

Now, let us calculate the number of mole of MgCl2. This is illustrated below:

Molarity Mass of MgCl2 = 24 + (2x35.5) = 24 + 71 = 95g/mol

Mass of MgCl2 = 151g

Number of mole of MgCl2 =?

Number of mole = Mass /Molar Mass

Number of mole of MgCl2 = 151/95

Number of mole of MgCl2 = 1.59mole

Now we can calculate the molarity of MgCl2 as follow:

Mole = 1.59mole

Volume = 1L

Molarity =?

Molarity = mole /Volume

Molarity = 1.59/1

Molarity = 1.59mol/L

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It represents a decomposition reaction because one reactant breaks apart and forms two products.

Explanation:

The type of chemical reaction depicted by Jane represents a decomposition reaction because on reactant breaks apart and forms two products.

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8 0
2 years ago
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Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c
Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>

<em />

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>

<em />

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M

6 0
2 years ago
in the early studies of chemistry, scientists used properties and changed to help identify compounds. this is still done today.
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2 years ago
What is the specific heat of an unknown substance if a 2.50 g sample releases 12 calories as its
pishuonlain [190]

Answer:

c = 4016.64 j/g.°C

Explanation:

Given data:

Mass of substance = 2.50 g

Calories release = 12 cal (12 ×4184 = 50208 j)

Initial temperature = 25°C

Final temperature = 20°C

Specific heat of substance = ?

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Solution:

Q = m.c. ΔT

ΔT  = T2 - T1

ΔT  = 20°C - 25°C

ΔT  = -5°C

50208 j = 2.50 g . c. -5°C

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7 0
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ladessa [460]

Answer:

The water cycle will affect the climate by causing rain and snow.

Explanation:

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4 0
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