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3 years ago

When matter heats up, dose the molecules close together?

2 answers:

8 0

Yes because when they heat up you sweat, and when they cool down they would probably separated because you are at your normal temperate

Dennis_Churaev [7]3 years ago

6 0

Ok don't look at the bottom answer, that's incorrect. When matter heats up, molecules start to separate. That's why gas has no physical form, unlike solids. Ice, for example, is solid, but when it heats up, the molecules strat to move apart and become a liquid and then a gas. Hope I helped!

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What is term redox reaction

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3 years ago

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Which of the following best describes exothermic chemical reactions?

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3 years ago

Is mud a suspension, solution, or colloid?

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2 years ago

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At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T

Anarel [89]

Answer: The freezing point of solution is -0.974°C

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M$

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

$0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (glucose) = 89.18 g

$M_{solute}$ = Molar mass of solute (glucose) = 180.16 g/mol

$W_{solvent}$ = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC$

Hence, the freezing point of solution is -0.974°C

8 0

3 years ago

What is the concentration of NaCI in an aqueous solution that contains 0.032 grams of NaCI in 600. Grams of the solution

jolli1 [7]

Answer:

0.00032 Grams of NaCl per 1 gram of the solution

Explanation:

8 0

3 years ago