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vampirchik [111]

3 years ago

13

What does water moves under earth's surface to fill in open spaces in rocks or soil?

1 answer:

Trava [24]3 years ago

4 0

Water moves under the earth's surface to fill in open spaces in rocks or soil. This type of water is called groundwater.

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The volume of 7.91 M HCl needed to make 196.1 mL of 2.13 M HCl is ____.

Alona [7]

Answer:

a. 52.8

Explanation:

To find the number of moles of HCl we use the relation M₁V₁=M₂V₂

where M₁ is the initial molarity, M₂ the new molarity, V₁ the initial volume used, and V₂ the final volume obtained.

M₁=7.91 M

M₂=2.13 M

V₁=?

V₂=196.1 mL

Replacing these values in the relationship.

M₁V₁=M₂V₂

7.91 M× V₁=2.13 M×196.1 mL

V₁=(2.13 M×196.1 mL)/7.91 M

=52.8 mL

3 0

3 years ago

Which statement is true? Postively charged objects attract other positively charged objects negatively charged objects attract o

ANEK [815]

Answer:

Explanation:one is true because if it is positiveit would not be attracting it would be sepperating

8 0

2 years ago

Which layer of rock contains more humus

sleet_krkn [62]

It is the horizon layer
~Hope this helps~

4 0

3 years ago

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

$CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)$

Calculate the amount of $CO_{2}$ dissolved as follows.

$CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}$

It is given that $K_{CO_{2}}$ = 0.032 M/atm and $P_{CO_{2}}$ = $1.9 \times 10^{-4}$ atm.

Hence, $[CO_{2}]$ will be calculated as follows.

$[CO_{2}]$ = $K_{CO_{2}} \times P_{CO_{2}}$

= $0.032 M/atm \times 1.9 \times 10^{-4}atm$

= $0.0608 \times 10^{-4}$

or, = $0.608 \times 10^{-5}$

It is given that $K_{a} = 4.46 \times 10^{-7}$

As, $K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}$

$4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}$

$[H^{+}]^{2}$ = $2.71 \times 10^{-12}$

$[H^{+}]$ = $1.64 \times 10^{-6}$

Since, we know that pH = $-log [H^{+}]$

So, pH = $-log (1.64 \times 10^{-6})$

= 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0

2 years ago

Why is it important to do further investigations to determine if a chemical change really occurred?

KatRina [158]

Answer:

x

Explanation:

5 0

2 years ago