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Varvara68 [4.7K]
3 years ago
15

The concentration of urea in a solution prepared by dissolving 16 g of urea in 20 g of H2OH2O is ________% by mass. The molar ma

ss of urea is 60.0 g/mol. The concentration of urea in a solution prepared by dissolving 16 g of urea in 20 g of is ________% by mass. The molar mass of urea is 60.0 g/mol. 44 80 0.80 0.48 0.44
Chemistry
1 answer:
Arada [10]3 years ago
4 0

Answer:

44

Explanation:

Given that :

Mass of solute = Mass of urea = 16g

Mass of water = 20g

Mass of solution = (mass of solute + mass of solvent) = (mass of urea + mass of water) = (16g + 20g) = 36g

Percentage Mass = (mass of solute / mass of solution) * 100%

Percentage Mass = (16 / 36) * 100%

Percentage Mass = 0.4444444 x 100%

Percentage Mass = 44.44%

Percentage Mass = 44%

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Calculate the mass of aluminum in 500 g of Al(C2H3O2)3
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Answer: 66.2 g

Explanation:

1) The ratio of Al in the molecule is  1 mol to 1 mol .

2) The mass of 1 mol of molecules of Al (CH2H3O2)3 is the molar mass of the compound.

3) You calculate the molar mass of the compound using the atomic masses of each atom, in this way:

Al: 27 g/mol
C: 2 * 3 * 12 g/mol = 72 g/mol
H: 3 * 3 * 1 g/mol = 9 g/mol
O: 2 * 3 * 16 g/mol = 96 g/mol

Molar mass = 27 g/mol + 72 g/mol + 9 g/mol + 96 g/mol = 204 g/mol

4) Set a proportion:

    27 g/mol                x
-------------------- =  ----------
   204 g/mol            500 g

5) Solve for x:

x = 500 g * 27 g/mol / 204 g/mol = 66.2 g
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3 years ago
What is the empirical formula
Radda [10]

Answer:

a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.

Explanation:

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2 years ago
How many moles of lithium are in 83 grams of lithium?
german
1 moles lithium or 6.941 grams
8 0
3 years ago
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Please help!! I was sick when he went over this
Alexus [3.1K]

Answer:

5.70×10^-11 m

Explanation:

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Determine the value of the equilibrium constant, Kgoal, for the reaction C(s)+12O2(g)+H2(g)⇌12CH3OH(g)+12CO(g), Kgoal=? by makin
Licemer1 [7]

Answer:

1.71x10²⁷

Explanation:

If we sum 1/2 of (3) + 1/2 of (1):

1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷  = 4.58x10²³

1/2 (1)   1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8

C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>

K' = 4.58x10²³ * 11.8 = 5.42x10²⁴

+1/2 (2):

<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2

C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)

K'' = 5.42x10²⁴* 316.2 =

<h3>1.71x10²⁷</h3>

5 0
3 years ago
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