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2 years ago

Increasing which factor will not increase the rate of a chemical reaction?

2 answers:

vladimir1956 [14]2 years ago

6 0

Two substances cannot react with each other unless their constituent particles (molecules, atoms, or icons) come into contact if there is no contact the reaction rate will be ZERO

Valentin [98]2 years ago

3 0

Answer:

c. product

i had the test

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Calculate the standard entropy of vaporization of ethanol, C2H5OH, at 285.0 K, given that the molar heat capacity at constant pr

MArishka [77]

Answer: The standard entropy of vaporization of ethanol is 0.275 J/K

Explanation:

$C_2H_5OH(l)\rightleftharpoons C_2H_5OH(g)$

Using Gibbs Helmholtz equation:

$\Delta G=\Delta H-T\Delta S$

For a phase change, the reaction remains in equilibrium, thus $\Delta G=0$

$\Delta H=T\Delta S$

Given: Temperature = 285.0 K

$\Delta H=78.3J/mol$

Putting the values in the equation:

$78.3J=285.0K\times \Delta S$

$\Delta S=0.275J/K$

Thus the standard entropy of vaporization of ethanol is 0.275 J/K

4 0

3 years ago

How does a tree get energy?

Snezhnost [94]

Your answer is B. It gets energy from the Sun.
Have a great day :)

6 0

2 years ago

Read 2 more answers

Calculate the freezing point and boiling point of a solution containing 8.15 g of ethylene glycol (C2H6O2) in 96.3 mL of ethanol

pishuonlain [190]

Answer: The freezing point of solution is -117.54°C and the boiling point of solution is 80.48°C

Explanation:

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

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

Density of ethanol = 0.789 g/mL

Volume of ethanol = 96.3 mL

Putting values in above equation, we get:

$0.789g/mL=\frac{\text{Mass of ethanol}}{96.3mL}\\\\\text{Mass of ethanol}=(0.789g/mL\times 96.3mL)=75.98g$

Calculating the freezing point:

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 = -114.1 °C

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

$K_f$ = molal freezing point elevation constant = 1.99°C/m

$m_{solute}$ = Given mass of solute (ethylene glycol) = 8.15 g

$M_{solute}$ = Molar mass of solute (ethylene glycol) = 62 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

$-114.1-\text{Freezing point of solution}=1\times 1.99^oC/m\times \frac{8.15\times 1000}{62g/mol\times 75.98}\\\\\text{Freezing point of solution}=-117.54^oC$

Hence, the freezing point of solution is -117.54°C

Calculating the boiling point:

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

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

Boiling point of pure solution = 78.4°C

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

$K_b$ = molal boiling point elevation constant = 1.20°C/m.g

$m_{solute}$ = Given mass of solute (ethylene glycol) = 8.15 g

$M_{solute}$ = Molar mass of solute (ethylene glycol) = 62 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-78.4=1\times 1.20^oC/m\times \frac{8.15\times 1000}{62\times 75.98}\\\\\text{Boiling point of solution}=80.48^oC$

Hence, the boiling point of solution is 80.48°C

3 0

3 years ago

The number of atoms in 2 moles of aluminium is if the 1mole is 6.022*10 the power of 23 or 10^23

patriot [66]

Answer:

1.204 × 10²⁴ atoms

Explanation:

According to this question, one mole of aluminum (Al) atom contains 6.02 × 10²³ atoms.

If two moles of aluminum are given, this means that there will be 2 × 6.02 × 10²³ atoms of aluminum

2 × 6.02 × 10²3

= 12.04 × 10²³

= 1.204 × 10²⁴ atoms.

3 0

3 years ago

Which of these contains information about the elements

Monica [59]

Post the question so we can help

7 0

3 years ago