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3 years ago

The density of a solid reactant affects the rate of a reaction. TRUE FALSE

Chemistry

2 answers:

levacccp [35]3 years ago

5 0

Answer:

The correct answer is "True".

Explanation:

The rate of a reaction is affected by multiple factors, including the density of a solid reactant. The higher the density of a solid reactant the higher the rate of a reaction will be. A higher density of reactant increases the probability that collisions take place, and the more collisions the more reactions will occur.

Shalnov [3]3 years ago

3 0

yes it is true it reàlly happens

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DONT SKIPP What type of bond will occur between lithium and bromine?

den301095 [7]

An ionic bond is formed between lithium and bromine.

4 0

3 years ago

A chemist measures the energy change ΔH during the following reaction: C3H8 (g) +5O2 (g) →3CO2 (g) +4H2O (l) =ΔH−2220.kJ Use the

statuscvo [17]

Answer:

The reaction is exothermic.

Yes, released.

The heat released is 4,08x10³ kJ.

Explanation:

For the reaction:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

The ΔH is -2220 kJ, As ΔH is <0, The reaction is exothermic.

As the reaction is exothermic, the heat of the reaction will be released.

The heat released in 81,0g is:

81,0g C₃H₈× $\frac{1mol}{44,1g}$ × $\frac{2220kJ}{1mol}$ = 4,08x10³ kJ

-Using molar mass of C₃H₈ to convert mass to moles and knowing that there are released 2220 kJ per mole of C₃H₈-

I hope it helps!

3 0

3 years ago

the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel

shtirl [24]

Answer: The entropy change of the process is $2.0\times 10^2J/K$

Explanation:

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=n\times \frac{\Delta H_{f}}{T}$

where,

$\Delta S$ = Entropy change

n = moles of acetone = 6.3 moles

$\Delta H_{f}$ = enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol (Conversion factor: 1 kJ = 1000 J)

T = temperature of the system = $-94.7^oC=[273-94.7]=178.3K$

Putting values in above equation, we get:

$\Delta S=\frac{6.3mol\times 5700J/mol}{178.3K}\\\\\Delta S=201.4J/K=2.0\times 10^2J/K$

Hence, the entropy change of the process is $2.0\times 10^2J/K$

4 0

3 years ago

55 L of a gas at 25oC has its temperature increased to 35oC. What is its new volume?

ladessa [460]

Answer:

Approximately 56.8 liters.

Assumption: this gas is an ideal gas, and this change in temperature is an isobaric process.

Explanation:

Assume that the gas here acts like an ideal gas. Assume that this process is isobaric (in other words, pressure on the gas stays the same.) By Charles's Law, the volume of an ideal gas is proportional to its absolute temperature when its pressure is constant. In other words

$\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1}$ ,