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sergey [27]
3 years ago
7

What must be true for a substance to be able to dissolve in water?

Chemistry
1 answer:
zvonat [6]3 years ago
3 0

In order for a solute to dissolve in a solvent, the attractive forces between solute particles and the solvent particles must be stronger than the attractive forces between solute-solute and solvent-solvent particles. This is important so that the solute will remain in solution.

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Choose all of the answers that describe beneficial microorganisms.
REY [17]

Answer:

1, 4,3,2

Explanation:

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2 years ago
What are weak bonds that allow flexibility in an enzyme
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Enzymes are characterized to have weak bonds because their tertiary structure could easily bend and break because it will have to adjust to the shape of the substrate. It could be done via induced fitting or lock-and-key theory. These weak bonds are intermolecular forces like the London forces, electrostatic interactions and hydrogen bonding.
7 0
3 years ago
4HCl(g) + Si (s) ⇌ SiCl4 (l) + 2H2 (g) would be classified as a(n):
zheka24 [161]

Answer:

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3 0
2 years ago
The freezing point of diet soda is higher than the freezing point of regular soda, but lower than 0 degrees celcius, the freezin
Zepler [3.9K]

Answer:

Explanation:

Both Diet Soda and regular soda contain sweeteners.

When a solute is dissolved in solution, the solution undergoes *freezing point depression* it freezing point reduces. The magnitude of freezing point depression is directly proportional to the amount of solute in a solution.

Since soda Both regular or diet soda contains more solute than water , their freezing point is will consequently be lower than water

4 0
3 years ago
The freezing point of benzene C6H6 is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is DDT . H
Vlad1618 [11]

Answer: 0.028 grams

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

\Delta T_f=k_f\times m

or,

\Delta T_f=k_f\times \frac{\text{ Mass of solute in g}\times 1000}{\text {Molar mass of solute}\times \text{ Mass of solvent in g}}

where,

\Delta T_f = change in freezing point

k_f = freezing point constant  (for benzene} =5.12^0Ckg/mol

m = molality

Putting in the values we get:

0.400^0C=5.12\times \frac{\text{ Mass of solute in g}\times 1000}{354.5\times 209.0}

{\text{ Mass of solute in g}}=0.028g

0.028 grams of DDT (solute) must be dissolved in 209.0 grams of benzene to reduce the freezing point by 0.400°C.

4 0
3 years ago
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