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kari74 [83]
3 years ago
13

½H2(g) + ½I2(g) → HI(g), ΔH = + 26 kJ/mole 117 kJ/mole + C(s) + 2S(s) → CS2(g) The temperature of the surroundings will:

Physics
2 answers:
Alona [7]3 years ago
8 0
From the above reaction the temperature of the surroundings will increase.
Alexus [3.1K]3 years ago
4 0

Answer:

Increase

Explanation:

The given equation is an example of an exothermic reaction. Hence heat is release in the surrounding and the temperature of the surrounding will increase.

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antoniya [11.8K]
I know it’s the Coulomb’s law and that I’m pretty sure the answer would be C.Inverse Square.
4 0
3 years ago
I need help with this answer
7nadin3 [17]

Answer:

double replacement

Explanation:

sorry if im wrong

8 0
3 years ago
Which of the following describes the normal shape of a DNA molecule?
Rudiy27

Answer:

A. Two strands of nucleotides bonded together at their bases,

twisting to form a double helix

Explanation:

Hope this helps! Can I have brainliest?

5 0
2 years ago
Five moles of an ideal monatomic gas with an initial temperature of 121 ∘C expand and, in the process, absorb an amount of heat
liq [111]

Answer:

  T₂ ≈ 107.85∘C

Explanation:

The question didn't state if the volume is constant or not as such, we can apply the first law of thermodynamic

From the first law of thermodynamic,

ΔU =  Q - W

where ΔU = Internal Energy, Q = Quantity of heat absorbed, W = Amount of work done.

Q = 1200 J and W = 2020 J

∴ ΔU = 1200 -2020 = -820 J.

Using the ideal gas equation,

ΔU = 3/2nRΔT...................................equation 1

where n = number of moles, R = Molar gas constant, ΔT = Change in temperature = (T₂ - T₁).

Modifying equation 1,

ΔU = 3/2nR(T₂ -T₁)...............................equation 2.

making T₂ the subject of the relation in equation 2,

T₂ =  {2/3(ΔU)/nR}+T₁........................ equation 3

where T₁=121∘C, R= 8.314 J / mol, n=5 moles, ΔU=-820 J

Substituting these values into equation 3,

∴ T₂ ={ 2/3(-820)/(5×8.314)}+121

   T₂ = {2×(-820)/ (3×5×8.314)}+121  

   T₂={-1640/124.71}+ 121

   T₂ = {-13.151} + 121

 ∴T₂  = 121 - 13.151 = 107. 849∘C

    T₂ ≈ 107.85∘C

3 0
3 years ago
An area experiences thunderstorms with high winds and a drop in temperature. Which weather event most likely occurred? A. A stat
gavmur [86]

Answer:

C a fast-moving cold front moved through the area.

Explanation:

This is because, since there is a there is a thunderstorm and high winds in the area, this can only be caused by a fast moving front. Also there is a temperature drop, this can only be caused by the fast moving cold front since a cold front has a low temperature.

Thus, for the area to experience thunderstorms with high winds and a drop in temperature, <u>a fast-moving cold front moved through the area.</u>

5 0
3 years ago
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