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maria [59]
3 years ago
11

When you boil water, why does the level of liquid decrease?

Chemistry
1 answer:
Ganezh [65]3 years ago
6 0

Answer:

when you boil water, you convert it into water vapor, which leaves the pot and mixes with the atmosphere. If you boil the pot long enough, eventually all the water in it is converted to vapor and leaves. the pot is then empty.

Explanation:

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What is the difference between conjugate acid-base pair?
NeX [460]

Answer:

B. a H+ ion is the answer dear.

Explanation:

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5 0
3 years ago
EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?
AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq).

The equilibrium constant for this reaction is called the self-ionization constant of water:

K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}].

Note that water isn't part of this constant.

The value of K_w at 25 °C is 10^{-14}. How to memorize this value?

  • The pH of pure water at 25 °C is 7.
  • [\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}
  • However, [\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3} for pure water.
  • As a result, K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14} at 25 °C.

Back to this question. [\text{H}^{+}] is given. 25 °C implies that K_w = 10^{-14}. As a result,

\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}.

8 0
3 years ago
Which ones react with water?
slavikrds [6]

Answer:

I think its the last one

Explanation:

its supposed to be H 2 O 2.

8 0
2 years ago
I have no idea how to do this help plz!
Wittaler [7]
The answes are
C)
Then it is
)A
Im doing this for class too bruh
8 0
3 years ago
Read 2 more answers
A sample of oxygen gas is compressed from 30.6 L to 1.8 L at constant temperature pressure of 1.8 atm. Calculate the amount of e
ad-work [718]

Answer:

the change in the internal energy of the system is 3,752.67 J

Explanation:

Given;

initial volume of the gas, V₁ = 30.6 L

final volume of the gas, V₂ = 1.8 L

constant pressure of the gas, P = 1.8 atm

Energy released by the system, Q = 1.5 kJ = 1,500 J

Apply pressure-volume work equation, to determine the work done on the gas;

w = -PΔV

w = -P(V₂ - V₁)

w = - 1.8 atm(1.8 L - 30.6 L)

w = 51.84 L.atm

w = 51.84 L.atm x 101.325 J/L.atm

w = 5,252.67 J

The change in the internal energy of the system is calculated as;

ΔU = Q + w

Since the heat is given out, Q = - 1,500 J

ΔU = -1,500 J  +  5,252.67 J

ΔU = 3,752.67 J

Therefore, the change in the internal energy of the system is 3,752.67 J

3 0
3 years ago
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