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suter [353]
3 years ago
8

Calculate the concentration of OH- in a solution that contains 3.9 x 10-4 M H3O+ at 25°C. Identify the solution as acidic, basic

or neutral. Calculate the concentration of OH- in a solution that contains 3.9 x 10-4 M H3O+ at 25°C. Identify the solution as acidic, basic or neutral. 3.9 × 10-4 M, neutral 2.7 × 10-2 M, basic 2.6 × 10-11 M, basic 2.7 × 10-2 M, acidic 2.6 × 10-11 M, acidic
Chemistry
1 answer:
aalyn [17]3 years ago
8 0

Answer:

[OH⁻] = 2,6x10⁻¹¹

Acidic

Explanation:

The kw in water is:

2 H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)

kw = [OH⁻] [H₃O⁺] = 1,00x10⁻¹⁴

If concentracion of H₃O⁺ is 3,9x10⁻⁴M:

[OH⁻] [3,9x10⁻⁴M] = 1,00x10⁻¹⁴

<em>[OH⁻] = 2,6x10⁻¹¹</em>

pH is defined as - log[H₃O⁺]. If pH>7,0 the solution is basic, if pH<7,0 solution is acidic, if pH=7,0 solution is neutral.

In this problem,

pH = - log [3,9x10⁻⁴M] = <em>3,4</em>

As pH is < 7.0, the solution is <em>acidic</em>

<em></em>

I hope it helps!

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Determine the oxidation number of sodium in Na202
Olegator [25]

Answer:

+1

Explanation:

Na₂O₂

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Thus, we can obtain the oxidation number of sodium (Na) in Na₂O₂ as illustrated below:

Na₂O₂ = 0 (oxidation number of ground state compound is zero)

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Collect like terms

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Divide both side by 2

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5 0
2 years ago
A sample of gas with an initial volume of 12.5 L at a pressure of 784 torr and a temperature of 295 K is compressed to a volume
Kipish [7]

Answer:

Final pressure in (atm) (P1) = 6.642 atm

Explanation:

Given:

Initial volume of gas (V) = 12.5 L

Pressure (P) = 784 torr

Temperature (T) = 295 K

Final volume (V1) = 2.04 L

Final temperature (T1) = 310 K

Find:

Final pressure in (atm) (P1) = ?

Computation:

According to combine gas law method:

\frac{PV}{T} =\frac{P1V1}{T1} \\\\\frac{(784)(12.5)}{295} =\frac{(P1)(2.04)}{310}\\\\33.22 = \frac{(P1)(2.04)}{310}\\\\P1=5,048.18877

⇒ Final pressure (P1) = 5,048.18877 torr

⇒ Final pressure in (atm) (P1) = 5,048.18877 torr / 760

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3 0
3 years ago
Based on the diagram below, how much of the excess reactant is left over? *
Alexxandr [17]

Answer:

3 of lunchmeat and 2 slices of cheese

Explanation:

From the question given,

Each sandwich contains:

2 bread + 3 lunch meat + 1 cheese.

Now, the limiting reactant is the slice of bread.

We can determine the leftover as follow:

1. For the lunchmeat.

From the simple equation above,

2 bread requires 3 lunchmeat.

Therefore, 6 bread will require = (6 x 3)/2 = 9 lunchmeat.

Lunchmeat given = 12

Lunchmeat required = 9

Leftover lunchmeat = 12 – 9 = 3

Therefore, 3 lunchmeat is leftover.

2. For cheese.

From the simple equation above,

2 bread requires 1 cheese.

Therefore, 6 bread will require = 6/2 = 3 cheese.

Cheese given = 5

Cheese required = 3

Leftover cheese = 5 – 3 = 2

Therefore, 2 cheese is leftover.

From the simple illustrations above,

3 lunchmeat and 2 cheese are leftover.

7 0
3 years ago
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