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cluponka [151]
4 years ago
6

At a certain temperature, the ph of a neutral solution is 7.43. what is the value of kw at that temperature? express your answer

numerically using two significant figures.
Chemistry
1 answer:
Delvig [45]4 years ago
7 0

The formula for Kw is:

Kw= [H]*[OH]

where [H] and [OH] are concentrations 

We are given the pH so we can calculate the [H] concentration from the formula:
pH=-log[H] 
7.43 = -log[H]

[H] = 3.715x10^-8 M

In neutral solution, [H] = [OH], therefore: 

<span>[OH]= 3.715x10^-8 M

Calculating Kw:

Kw = (3.715x10^-8)*( 3.715x10^-8)</span>

<span>Kw = 1.38 x 10^-15 </span>

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The teacher in item 10 also needs to order plastic tubing. If each of the 60 students need 750 mm of tubing, what length of tubi
Alex777 [14]

45 m. If each student needs 750 mm of tubing, the teacher should order 45 m of tubing.

a) Find the <em>length in millimetres</em>

Length = 60 students x (750 mm tubing/1 student) = 45 000 mm tubing

b) Convert <em>millimetres to metres </em>

Length = 45 000 mm tubing x (1 m tubing/1000 mm tubing) = 45 m tubing

6 0
3 years ago
Which of the following compounds has only one atom of carbon?
frez [133]
<h3>Answer: CO2</h3>

Explanation:

Well C2H6 there are 2 carbon and 6 hydrogen atoms

and having no number on the right of the letter means there is only one atom

8 0
4 years ago
GeF 3H is formed from GeH 4 and GeF 4 in the combination reaction: GeH 4 3GeF 4 4GeF 3H If the reaction yield is 92.6%, how many
Olin [163]

Answer:

6.48

Explanation:

GeH4 + 3 GeF4 > 4 GeF3H

3 moles of GeF4 react to form 4 moles of GeF4

Therefore, (8 x 3/4)/0.926

= 6.48 moles

4 0
4 years ago
Calculate the pH of each solution at 25∘C
JulijaS [17]

Answer: pH of HCl =5, HNO3 = 1,

NaOH = 9, KOH = 12

Explanation:

pH = -log [H+ ]

1. 1.0 x 10^-5 M HCl

pH = - log (1.0 x 10^-5)

= 5 - log 1 = 5

2. 0.1 M HNO3

pH = - log (1.0 x 10 ^ -1)

pH = 1 - log 1 = 1

3. 1.0 x 10^-5 NaOH

pOH = - log (1.0 x 10^-5)

pOH = 5 - log 1 = 5

pH + pOH = 14

Therefore , pH = 14 - 5 = 9

4. 0.01 M KOH

pOH = - log ( 1.0 x 10^ -2)

= 2 - log 1 = 2

pH + pOH = 14

Therefore, pH = 14 - 2 = 12

8 0
4 years ago
Calculate the volume of carbon dioxide at 20.0°C and 0.941 atm produced from the complete combustion of 4.00 kg of methane. Comp
tankabanditka [31]

Answer:

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6390.89 Liters.

More volume of carbon dioxide gas was released on combustion of 4.00 kg of propane in comparison to the volume of carbon dioxide released on combustion of 4.00 kg of methane.

Explanation:

Methane

CH_4+2O_2\rightarrow CO_2+2H_2O

Mass of methane = 4.00 kg = 4000 g (1 kg = 1000 g)

Moles of methane = \frac{4000 g}{16 g/mol}=250 mol

According to reaction, 1 mole of methane gives 1 mole of carbon dioxide gas,then 250 moles of methane will give :

\frac{1}{1}\times 250 mol=250 mol of carbon dioxide gas

Moles of carbon dioxide gas = n = 250 mol

Pressure of carbon dioxide gas = P = 0.941 atm

Temperature of carbon dioxide gas = T = 20.0°C = 20.0+273 K = 293 K

Volume of carbon dioxide gas = V

PV=nRT (Ideal gas equation)

V=\frac{nRT}{P}=\frac{250 mol\times 0.0821 atm L/mol K\times 293 K}{0.941 atm}=6390.89 L

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6390.89 Liters.

Propane

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

Mass of propane = 4.00 kg = 4000 g (1 kg = 1000 g)

Moles of propane = \frac{4000 g}{44 g/mol}=90.91 mol

According to reaction, 1 mole of propane gives 3 mole of carbon dioxide gas,then 90.91 moles of propane will give :

\frac{3}{1}\times 90.91 mol=272.73 mol of carbon dioxide gas

Moles of carbon dioxide gas = n = 272.73 mol

Pressure of carbon dioxide gas = P = 0.941 atm

Temperature of carbon dioxide gas = T = 20.0°C = 20.0+273 K = 293 K

Volume of carbon dioxide gas = V

PV=nRT (Ideal gas equation)

V=\frac{nRT}{P}=\frac{272.73 mol\times 0.0821 atm L/mol K\times 293 K}{0.941 atm}=6,971.95 L

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6,971.95 Liters.

Volume of carbon dioxide given by combustion of 4.00 kg of methane = 6390.89 Liters.

Volume of carbon dioxide given by combustion of 4.00 kg of propane = 6,971.95 Liters.

6390.89 Liters < 6,971.95 Liters

More volume of carbon dioxide gas was released on combustion of 4.00 kg of propane in comparison to the volume of carbon dioxide released on combustion of 4.00 kg of methane.

7 0
3 years ago
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