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Gnom [1K]
3 years ago
8

If a buffer solution is 0.190 m in a weak acid (ka = 8.2 × 10-5) and 0.590 m in its conjugate base, what is the ph?\

Chemistry
1 answer:
Lubov Fominskaja [6]3 years ago
5 0
<span>You use the Henderson - Hasselbalch equation pH = pKa + log ([salt]/[acid]) pKa = -log (8.2*10^-5) = 4.081 pH = 4.081 + (0.590/0.190) pH = 4.081 + log 3.105 pH = 4.081 + 0.49206 pH = 4.573</span>
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If a climate scientist says, "The CO2 concentration in 2018 was 407.4 ppm." What<br> do they mean? *
Alinara [238K]

Answer:

Ppm <u>means the concentration of vapors or gases</u> expressed in parts per million of polluted air.

Explanation:

In Chemistry and Physics, part per million or PPM is a measure that expresses the number of units of a certain existing substance for every million units of the mixture. (Concentration)

In climate the equation is used so:

  • ppm = parts of polluted air / million parts of air .
  • ppm = Liters of polluted air / 10∧6 Liters  of air.

This is used to find the concentration of gases expressed as parts of gas per million parts of the

 polluted air.

In this case, it would be<u> 407.4L of CO2 per 10∧6L of air.</u>

6 0
3 years ago
) determine the theoretical yield and the percent yield if 21.8 g of k2co3 is produced from reacting 27.9 g ko2 with 29.0 l of c
Airida [17]
The Balanced Chemical Equation is as follow;

                         4 KO₂  +  2 CO₂    →    2 K₂CO₃  +  3 O₂

First find out the Limiting Reagent,
According to equation,

         284 g (4 moles) KO₂ reacted with  =  44.8 L (2 moles) of CO₂
So,
                  27.9 g of KO₂ will react with  =  X  L of CO₂


Solving for X,
                          X  =  (44.8 L × 27.9 g) ÷ 284 g

                          X  =  4.40 L of CO₂

Hence, to consume 27.9 g of KO₂ only 4.40 L CO₂ is required, while, we are provided with 29 L of CO₂, it means CO₂ is in excess and KO₂ is is limited amount, Therefore, KO₂ will control the yield of K₂CO₃. So,

According to eq.

         284 g (4 moles) KO₂ formed  =  138.2 g of K₂CO₃
So,
         27.9 g of KO₂ will form  =  X g of K₂CO₃

Solving for X,
                        X  =  (138.2 g × 27.9 g) ÷ 284 g

                        X  =  13.57 g of K₂CO₃

So, 13.57 g of K₂CO₃ formed is the theoretical yield.

%age Yield  =  13.57 / 21.8 × 100

%age Yield  =  62.24 %
3 0
4 years ago
Read 2 more answers
Exercise 2:
vagabundo [1.1K]

Answer:

1. KCLO3------>KCL + 3/2O2(g)

2. 122.5g/mol

3. 0.2mol

4. 18.5g

7 0
3 years ago
Volume increases, pressure stays the same, temperature
Mila [183]
Temperature decreases (?)
4 0
3 years ago
How many formula units are there in 450 g of Na2S04?
Mashutka [201]

Formula units in 450 g of Na_{2} So_{4} is 1.93 × 10²⁴ formula units.

<u>Explanation:</u>

First we have to find the number of moles in the given mass by dividing the mass by its molar mass as,

$\frac{450 g}{142.04 g/mol} =  3.2 moles

Now, we have to multiply the number of moles of Na₂SO₄ by the Avogadro's number, 6.022 × 10²³ formula units/mol, so we will get the number of formula units present in the given mass of the compound.

3.2 mol × 6.022 × 10²³ = 1.93 × 10²⁴ formula units.

So, 1.93 × 10²⁴ formula units is present in 450g of Na₂SO₄.

4 0
3 years ago
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