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Hatshy [7]
3 years ago
5

What happens when you increase the temperature of a reaction?

Chemistry
2 answers:
aalyn [17]3 years ago
7 0
D. More collisions occur and the time required for the reaction decreases
This happens because according to collision theory, when energy (in this case, thermal energy) is applied to particles, they move/vibrate more quickly. 
Julli [10]3 years ago
7 0

Answer: Option (D) is the correct answer.

more collisions occur and the time required for the reaction decreases.

Explanation:

When we increase the temperature of reactant molecules then there will occur an increase in the kinetic energy of molecules.

Also,      K.E = \frac{3}{2}kT

So, kinetic energy is directly proportional to the temperature. As a result, more number of collisions will take place between the molecules due to which there will be an increase in the rate of reaction.

Therefore, we can conclude that when you increase the temperature of a reaction then more collisions occur and the time required for the reaction decreases.

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False it’s used in a lot of other countries
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A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin
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Answer : The mass of the water in two significant figures is, 3.0\times 10^1g

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron metal = 0.45J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

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m_2 = mass of water = ?

T_f = final temperature of mixture = 59.2^oC

T_1 = initial temperature of iron metal = 21.9^oC

T_2 = initial temperature of water = 63.5^oC

Now put all the given values in the above formula, we get

32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC

m_2=30.16g\approx 3.0\times 10^1g

Therefore, the mass of the water in two significant figures is, 3.0\times 10^1g

3 0
3 years ago
Calculate the energy required to heat 566.0mg of graphite from 5.2°C to 23.2°C. Assume the specific heat capacity of graphite un
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7.23 J

Explanation:

Step 1: Given data

  • Mass of graphite (m): 566.0 mg
  • Initial temperature: 5.2 °C
  • Final temperature: 23.2 °C
  • Specific heat capacity of graphite (c): 0.710J·g⁻¹K⁻¹

Step 2: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)

Q = 7.23 J

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We add 1.4 moles to 700 mL of water. What is the molarity
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