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fiasKO [112]
3 years ago
10

If the [H+] is 3.5x10-4, what is the [OH-]?

Chemistry
1 answer:
slava [35]3 years ago
8 0

Answer:

2.90 x 10⁻¹¹moldm⁻³

Explanation:

Given parameters:

[H⁺]  = 3.5 x 10⁻⁴mol/dm³

Unknown

[OH⁻]  = ?

Solution;

The ionic product of water can be used to solve this problem. It has been experimentally determined to be  1 x 10⁻¹⁴mol² dm⁻⁶

             [H⁺] [OH⁻]  = 1 x 10⁻¹⁴

Therefore;

             [OH⁻]   = \frac{1  x 10^{-14} }{H}    = \frac{1  x  10^{-14} }{3.5 x 10^{-4} }   = 0.29 x 10⁻¹⁰moldm⁻³

                                                            = 2.90 x 10⁻¹¹moldm⁻³

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Convert the pressure 525.4 torr to kPa.
Virty [35]

Answer:

A. 70.0 kpa

Explanation:

Given

Pressure = 525.4 torr

Required

Convert to kPa

From standard unit of conversion;

1\ torr = 0.133\ kPa

Multiply both sides by 525.4

525.4 * 1\ torr = 525.4 * 0.133\ kpa

525.4\ torr = 70.047572\ kpa

525.4\ torr = 70.0\ kpa <em>(Approximated)</em>

<em>From the list of given options, option A is the closest answer after 525.4 torr is converted to its kpa equivalent</em>

8 0
3 years ago
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How many H atoms are in 6.43g x 10^24 molecules of water
kow [346]

Answer:

12.86\times 10^{24} atoms of hydrogen

Explanation:

The molecular formula for a water molecule is H_2O.

From the above formula, we can conclude that 1 molecule of water contains 2 atoms of hydrogen and 1 atom of oxygen.

There are 6.43\times 10^{24} molecules of water.

1 molecule of water = 2 atoms of hydrogen.

∴ 6.43\times 10^{24} molecules of water = 2\times 6.43\times 10^{24}=12.86\times 10^{24} atoms of hydrogen.

Therefore, there are 12.86\times 10^{24} atoms of hydrogen in 6.43\times 10^{24} molecules of water.

5 0
3 years ago
How many liters are equivalent to 645 mililiters
abruzzese [7]

Answer:

0.645 liters

Explanation:

THE QUESTION IS equivalent 0.645 Liters

6 0
4 years ago
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Which of the following atoms forms an ionic bond with a sulfur atom?
Luda [366]
Well, depending on the charge, it could be Cu; if it has a charge of 2+
5 0
3 years ago
For the following galvanic cell, represented in line notation, determine what balanced half-reactions occur at each electrode. (
Alex787 [66]

Answer:

Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻

Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

Explanation:

Let's consider the notation of a galvanic cell.

Cr(s) | Cr³⁺(aq) || Ag⁺(aq) | Ag(s)

On the left, it is represented the anode (oxidation) and on the right, it is represented the cathode (reduction).

The half-reactions are:

Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻

Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

To have the global reaction, we have to multiply the reduction by 3 (so the number of electrons gained and lost are the same) and add both half-reactions.

Global reaction: Cr(s) + 3 Ag⁺(aq) ⇒ Cr³⁺(aq) + 3 Ag(s)

6 0
3 years ago
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