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Kobotan [32]
3 years ago
12

Use the collision theory to discuss how temp., concentration and surface area affect the rate of a chemical reaction?

Chemistry
1 answer:
pogonyaev3 years ago
4 0


The collision theory explains how chemical reactions are made and different reactions have different reaction rates

Hope it helps.

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Sam believes the rock is more denser than the pencil. He measured the mass of the rock to be 8.5 grams and volume to be 4.5mL. H
mezya [45]

Answer:

Rock

Explanation:

Let's calculate the density of each object:

Rock:

Density = \frac{mass}{volume}=\frac{8.5\ g}{4.5\ mL}=1.9\ g/mL

Pencil:

Density = \frac{mass}{volume}=\frac{2.4\ g}{4.5\ mL}=0.53\ g/mL

Therefore the rock is denser.

6 0
2 years ago
What is the mass of 975 mL of mercury? Its density is 13.5 g/mL
miss Akunina [59]

Answer:

72.22 g

Explanation:

975 mL Mercury× 13.5 g/mL = 72.22 g

7 0
3 years ago
What is the molarity of a KOH solution if it requires 20 milliters of 2.0m HCL to exactcly neutalize 20 milliters of the KOH sol
soldier1979 [14.2K]
HCL 2.0x 3.6790= KOH SOLITION
6 0
3 years ago
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Bryan and Alyssa are playing tug-of-war against Daniel and Maria. Bryan and Alyssa are exerting a combined force of 10 N. Daniel
mr_godi [17]

Answer:

D

Explanation:

4 0
3 years ago
Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.
Marianna [84]

The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
  • Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
  • Initial temperature (T₁) = –25 °C
  • Final temperature (T₂) = 0 °
  • Change in temperature (ΔT) = 0 – (–38) = 38 °C
  • Specific heat capacity (C) = 2050 J/(kg·°C)
  • Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
  • Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
  • Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 100 °C
  • Final temperature (T₂) = 160 °C
  • Change in temperature (ΔT) = 160 – 100 = 60 °C
  • Specific heat capacity (C) = 1996 J/(kg·°C)
  • Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
  • Heat for –38 °C to 0°C (Q₁) = 31160 J
  • Heat for melting (Q₂) = 133600 J
  • Heat for 0 °C to 100 °C (Q₃) = 167200 J
  • Heat for vaporization (Q₄) = 904000 J
  • Heat for 100 °C to 160 °C (Q₅) = 47904 J
  • Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

brainly.com/question/10286596

#SPJ1

7 0
2 years ago
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