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Sindrei [870]
3 years ago
5

The number of particles in one mole of a substance is:

Chemistry
1 answer:
spayn [35]3 years ago
3 0
Avogadros number is the answer
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How much 3.0 M NaOH is needed to neutralize 30. mL of 0.75 M H2SO4?
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15.0 mL

Explanation:

Please see the step-by-step solution in the picture attached below.

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A chemical reaction was carried out three times. The mass of the product was 8.93 g for the first trial, 8.94 g for the second t
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The three mass value measure are precise mass

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3 years ago
(06.01 MC)
skelet666 [1.2K]

Answer:

Part 1. When the balloon is filled half of the way, and placed into the freezer, it will shrink. This happens because kinetic molecular theory tells us that a decrease in temperature decreases the kinetic energy of the gas molecules in the balloon. Viscous gases like hydrogen are less likely to shrink.

Part 2. When the balloon is placed out in the hot sun, most likely the balloon will swell and grow. This happens because the kinetic energy of the gas molecules increases due to solar radiation transforming into heat energy and then transforming into kinetic energy. Sticky gases like neon are more likely to grow.

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7 0
3 years ago
The equilibrium of 2H 2 O(g) 2H 2 (g) + O 2 (g) at 2,000 K has a Keq value of 5.31 x 10-10. What is the Keq expression for this
nevsk [136]

Answer:

5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}

Explanation:

For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.

For a hypothetical reaction:

xA + yB ⇄ zC

The equilibrium constant is :

Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }

The given reaction involves the decomposition of H2O into H2 and O2

2H_{2}O\rightleftharpoons 2H_{2} + O_{2}

The equilibrium constant is expressed as :

Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}

Since Keq = 5.31*10^-10

5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}

3 0
3 years ago
Read 2 more answers
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