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2 years ago

I need helppppp!!!!!!!

Chemistry

1 answer:

Gala2k [10]2 years ago

7 0

Answer:

1. Endothermic

2. Exothermic

3. Exothermic

4. Endothermic

Explanation:

Just the laws of thermodynamics. I always like to remember exothermic as heat exiting or releasing by the phrase "exo" for exit.

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The Kelvin temperature required for 0.0470 mol of helium gas to fill a balloon to 1.20 L under 0.878 atm is

4vir4ik [10]

From the ideal gas law, PV = nRT, we can rearrange the equation to solve for T given the other parameters.

T = PV/nR

where P = 0.878 atm, V = 1.20 L, n = 0.0470 moles, and R = 0.082057 L•atm/mol•K. Plugging in our values, we obtain the temperature in Kelvin:

T = (0.878 atm)(1.20 L)/(0.0470 mol)(0.082057 L•atm/mol•K)
T = 273 K

So, the second answer choice would be correct.

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3 years ago

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Answer:

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2 years ago

Consider the equilibrium reaction and its equilibrium constant expression. Br 2 ( g ) + 2 NO ( g ) − ⇀ ↽ − 2 NOBr ( g ) K = [ NO

zavuch27 [327]

Answer:

$K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}$

Explanation:

Hello,

In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

$2 Br_2 ( g ) + 4 NO ( g ) \rightleftharpoons 4NOBr ( g )$

The suitable equilibrium constant turns out:

$K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}$

Or in terms of the initial equilibrium constant:

$K_2=K_1^2$

Since the second reaction is a doubled version of the first one.

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