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ladessa [460]
3 years ago
11

A sample of lemon juice has a pH of 3. What is the [OH-] concentration of the lemon juice? A) 11 B) 1 X 10+3 C) 1 X 10-3 D) 1 X

10-11
Chemistry
2 answers:
Likurg_2 [28]3 years ago
5 0

pH scale is used to determine the acidity of a solution

pH is calculated using the H⁺ concentration

pOH scale is used to determine how basic a solution is

pOH is calculated using the OH⁻ concentration

pH and pOH are related in the following formula

pH + pOH = 14

pH = 3

pOH = 14 - 3

pOH = 11

pOH is calculated using the following formula

pOH = - log[OH⁻]

[OH⁻] = antilog(-pOH)

[OH⁻] = antilog(-11)

[OH⁻] = 1 x 10⁻¹¹ M

the [OH⁻] concentration of lemon juince is D) 1 x 10⁻¹¹ M

eimsori [14]3 years ago
4 0
PH+pOH=14
pOH=14-pH=14-3=11
pOH=-log[OH⁻]
11=-log[OH⁻]
[OH⁻]= 10⁻¹¹ =1*10⁻¹¹ 
Answer is D.
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The answer for the following problem is mentioned below.

  • <u><em>Therefore the final temperature of the gas is 740 K</em></u>

Explanation:

Given:

Initial pressure of the gas (P_{1}) = 1.8 atm

Final pressure of the gas (P_{2})  = 4 atm

Initial temperature of the gas (T_{1}) = 60°C = 60 + 273 = 333 K

To solve:

Final temperature of the gas (T_{2})

We know;

From the ideal gas equation;

we know;

P  × V = n × R × T

So;

we can tell from the above equation;

 <u>   P ∝ T</u>

(i.e.)

      <em> </em>\frac{P}{T}<em> = constant</em>

        \frac{P_{1} }{P_{2} } = \frac{T_{1} }{T_{2} }

Where;

P_{1}  = initial pressure of a gas

P_{2} = final pressure of a gas

T_{1} = initial temperature of a gas

T_{2} = final temperature of a gas

        \frac{1.8}{4} = \frac{333}{T_{2} }

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8 0
3 years ago
0.5 gm of mixture of NH4Cl and NaCl was boiled with 25 ml of 0.95 N NaOH in a vessel till all the ammonia is expelled.The residu
GalinKa [24]

Answer:

72.66%

Explanation:

NH₄Cl reacts in presence of NaOH producing ammonia, NH₃, as follows:

NH₄Cl + NaOH → NaCl + NH₃ + H₂O

The residual NaOH reacts with H₂SO₄ as follows:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.

The equivalent-gram of H₂SO₄ are:

16mL * 0.1N * 1.06 = 1.696mEq.

As the complete residual solution is 100mL but the neutralization was made only with 10mL, the mEq you need to neutralize the residual NaOH is:

1.696mEq * (100mL / 10mL) = 16.96mEq.

The mEq of NaOH you add in the first are:

25mL * 0.95mEq = 23.75mEq

That means the NaOH that reacts = moles of NH₄Cl is:

23.75mEq - 16.96mEq = 6.79mEq = 6.79mmoles NH₄Cl =

6.79x10⁻³ moles NH₄Cl

In grams (Using molar mass NH₄Cl = 53.5g/mol):

6.79x10⁻³ moles NH₄Cl * (53.5g / mol) =

0.3633g of NH₄Cl are in the original mixture.

% is:

0.3633g/ 0.5g * 100 = 72.66%

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4 years ago
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