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Vlada [557]
3 years ago
6

The number of electrons in a neutral atom is equal to the number of protons? True or false

Chemistry
1 answer:
NNADVOKAT [17]3 years ago
3 0

Answer:

true

Explanation:

in order for the atom to be neutral the number of protons and electrons have to be the same.

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Which of these is something whose construction MOST LIKELY would be determined by the government? A) An airport B) A golf club C
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Describe an experiment to show that pressure acts in all directions in liquids.​
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<h3>Liquid's pressure</h3>

Since both liquids and gases may flow, they are both referred to as fluids. Fluids under rest pressure behave uniformly in all directions.

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<h3>Pressure and depth in liquids</h3>

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3 0
2 years ago
in general, as the temperature increases, the solubility of gases in water ____ and the solubility of most solids in water___
saul85 [17]

Answer:

in general, as the temperature increases, the solubility of gases in water <u>decreases</u> and the solubility of most solids in water <u>increases</u>.

4 0
3 years ago
What are the concentrations of A , A, B , B, and C C at equilibrium if, at the beginning of the reaction, their concentrations a
Elenna [48]

The question is incomplete, here is the complete question:

A reaction

A+B\rightleftharpoons C

has a standard free-energy change of -4.88 kJ/mol at 25°C

What are the concentrations of A, B, and C at equilibrium if at the beginning of the reaction their concentrations are 0.30 M, 0.40 M and 0 M respectively?

<u>Answer:</u> The equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.

<u>Explanation:</u>

Relation between standard Gibbs free energy and equilibrium constant follows:

\Delta G^o=-RT\ln K_c

where,

\Delta G^o = Standard Gibbs free energy = -4.88 kJ/mol = -4880 J/mol  (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314J/K mol

T = temperature = 25^oC=[273+25]K=298K

K_c = equilibrium constant of the reaction

Putting values in above equation, we get:

-4880J/mol=-(8.3145J/Kmol)\times 298K\times \ln K_c\\\\K_c=7.17

We are given:

Initial concentration of A = 0.30 M

Initial concentration of B = 0.40 M

Initial concentration of C = 0 M

The chemical reaction follows:

                               A+B\rightleftharpoons C

<u>Initial:</u>                 0.30  0.40      0

<u>At eqllm:</u>         0.30-x   0.40-x    x

The expression of equilibrium constant for the above reaction follows:

K_c=\frac{[C]}{[A][B]}

We are given:

K_c=7.17

Putting values in above equation, we get:

7.17=\frac{x}{(0.30-x)\times (0.40-x)}\\\\x=0.183,0.657

Neglecting the value of x = 0.657, because change cannot be greater than the initial concentration

So, equilibrium concentration of A = (0.30-x)=(0.30-0.183)=0.117M

Equilibrium concentration of B = (0.40-x)=(0.40-0.183)=0.217M

Equilibrium concentration of C = x=0.183M

Hence, the equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.

5 0
3 years ago
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