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shutvik [7]
3 years ago
5

How are mixtures and solutions different from each other?

Chemistry
1 answer:
lorasvet [3.4K]3 years ago
5 0
D. mixtures are easily separated while the solutions are not.
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Describe the difference between pure substance and mixtures ( give an example each)
kondaur [170]

Answer:

a pure substance consists only of one element or one compound

a mixture consists of two or more different substances, not chemically joined together.

examples:

pure substance : Hydrogen gas - Diamond - Gold metal.

mixture : water and oil - mixtures of sand and water - trail mix

Explanation:

4 0
2 years ago
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Why does a scientist form a hypothesis?
Kryger [21]
Answer: B (to provide a statement that can be tested with an experiment
8 0
3 years ago
Uv gel enhancements rely on ingredients from the monomer liquid and ________ chemical family.
pishuonlain [190]
<span>UV gel enhancements rely on ingredients from the monomer liquid and polymer powder chemical family. The chemicals from the polymer powder family </span><span>can absorb and retain extremely large amounts of a liquid relative to their own mass. Water-absorbing polymers, </span>can absorb aqueous solutions through hydrogen bonding with water molecules.




6 0
3 years ago
A certain reaction with an activation energy of 205 kj/mol was run at 485 k and again at 505 k . what is the ratio of f at the h
lapo4ka [179]
Arrhenius' Law relates activation energy, Ea, rate constant, K, and temperature, T as per this equation:

K (T) = A * e ^ (-Ea / RT), where R is the universal constant of gases and A is a constant which accounts for collision frequency..

Then you can find the ration between K's at two different temperatures as:

K1 = A * e ^ (-Ea / RT1)

K2 = A* e ^(-Ea / RT2)

=> K1 / K2 = e ^ { (-Ea / RT1) - Ea / RT2) }

=> K1 / K2 = e ^ {(-Ea/ R ) *( 1 / T1 - 1 T2) }

=> K1 / K2 = e^ { (-205,000 j/mol / 8.314 j/mol*k )* ( 1 / 505K - 1/ 485K) }

=> K1 / K2 = e ^ (2.0134494) ≈ 7.5

Answer: 7.5




8 0
3 years ago
HELP. I DONT KNOW WHAT TO DO. Barium sulfate, BaSO4, is made by the following reaction.
Ghella [55]

Answer:

\boxed{\text{66.95 g BaSO$_{4}$}}

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:         261.34                         233.39

              Ba(NO₃)₂ + Na₂SO₄ ⟶ BaSO₄ + 2NaNO₃

m/g:         75.00

1. Moles of Ba(NO₃)₂

\text{Moles of Ba(NO$_{3})_{2}$} = \text{75.00 g} \times \dfrac{\text{1 mol}}{\text{261.34 g}} = \text{0.286 98 mol}

2. Moles of BaSO₄

The molar ratio is (1 mol BaSO₄/1 mol Ba(NO₃)₂

\text{Moles of BaSO$_{4}$}= \text{0.286 98 mol Ba(NO$_{3})_{2}$ } \times \dfrac{\text{1 mol BaSO$_{4}$}}{\text{1 mol Ba(NO$_{3})_{2}$}} = \text{0.286 98 mol BaSO$_{4}$}

3. Mass of BaSO₄

\text{Mass of BaSO$_{4}$} = \text{0.286 98 mol BaSO$_{4}$} \times \dfrac{\text{233.39 g BaSO$_{4}$}}{\text{1 mol BaSO$_{4}$}} = \textbf{66.98 g BaSO$_{4}$}\\\\\text{The theoretical yield of barium sulfate is } \boxed{\textbf{66.98 g BaSO$_{4}$}}

8 0
3 years ago
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