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Kruka [31]
3 years ago
8

True or false there are equal numbers of protons and electrons in neutral atoms.

Chemistry
2 answers:
Romashka [77]3 years ago
6 0

Answer:

true

Explanation:

Akimi4 [234]3 years ago
5 0

true the answer is true



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1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?
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Hello!

1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

  • <u><em>We have the following data:</em></u>

Vo (initial volume) = 1.00 L  

V (final volume) = 473 mL → 0.473 L  

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)  

P (final pressure) = ? (in atm)

  • <u><em>We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:</em></u>

P_0*V_0 = P*V

1*1 = P*0.473

1 = 0.473\:P

0.473\:P = 1

P = \dfrac{1}{0.473}

\boxed{\boxed{P \approx 2.11\:atm}}\:\:\:\:\:\:\bf\green{\checkmark}

<u><em>Answer:  </em></u>

<u><em>The new pressure of the gas is 2.11 atm  </em></u>

___________________________________

\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

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Examine the statement about alchemy
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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}}

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