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rjkz [21]
3 years ago
13

What is the Ph of a 3.9* 10^-8 M OH- solution

Chemistry
1 answer:
Maksim231197 [3]3 years ago
3 0
Molarity = 3.9 * 10 ^ -8

pH = - log (3.9 * 10 ^ -8)

pH = - (-7.40893539)

pH = 7.409 = 7.4 ... Ans
You might be interested in
What is the pressure of 0.60 moles of a gas if its volume is 10.0 liters at 35.0°C
eduard

Answer: 1.52 atm

Explanation:

Given that:

Volume of gas V = 10.0L

Temperature T = 35.0°C

Convert Celsius to Kelvin

(35.0°C + 273 = 308K)

Pressure P = ?

Number of moles = 0.6 moles

Molar gas constant R is a constant with a value of 0.0821 atm L K-1 mol-1

Then, apply ideal gas equation

pV = nRT

p x 10.0L = 0.6 moles x (0.0821 atm L K-1 mol-1 x 308K)

p x 10.0L = 15.17 atm L

p = 15.17 atm L / 10.0L

p = 1.517 atm (round to the nearest hundredth as 1.52 atm)

Thus, the pressure of the gas is 1.52 atm

3 0
3 years ago
All scientific discoveries are made in a<br> laboratory.<br> True<br> False<br> pls helppp
goblinko [34]

Answer:

i would say false.

Explanation:

some discoveries are made as people continue exploring the world...........they are just prooved in the laboratory, while there are other which are still made in the lab and are also examined there.

4 0
3 years ago
Please answer this. List the three most abundant minerals in this bottle of mineral water.​
Lisa [10]
The three most abundant minerals in a water bottle are calcium, potassium, and magnesium. {etc}
6 0
3 years ago
What is the measure of AC?<br>5 units<br>13 units<br>26 units<br>39 units<br>​
Natalka [10]

Answer:

26 units

Explanation:

6 0
3 years ago
Read 2 more answers
Calculate Delta G for each reaction using Delta Gf values: answer kJ ...thank you
Leni [432]

Answer:

a) \Delta G=2.6kJ

b) \Delta G=-979.57kJ

c) \Delta G=264.21kJ

Explanation:

Hello,

In this case, in each reaction we must subtract the Gibbs free energy of formation the reactants to the Gibbs free energy of formation of the products considering each species stoichiometric coefficients. In such a way, the Gibbs free energy of formations are:

\Delta _fG_{H_2}=\Delta _fG_{I_2}=0kJ/mol\\\Delta _fG_{HI}=1.3kJ/mol\\\Delta _fG_{CO_2}=-394.4kJ/mol\\\Delta _fG_{CO}=-137.3 kJ/mol\\\Delta _fG_{NH_3}=16.7 kJ/mol\\\Delta _fG_{HCl}=-95.3kJ/mol\\\Delta _fG_{MnO_2}=465.37kJ/mol\\\Delta _fG_{Mn}=0kJ/mol\\\Delta _fG_{NH_4Cl}=-342.81kJ/mol

So we proceed as follows:

a)

\Delta G=2\Delta _fG_{HI}-\Delta _fG_{H_2}-\Delta _fG_{I_2}\\\\\Delta G=2*1.3\\\\\Delta G=2.6kJ

b)

\Delta G=\Delta _fG_{Mn}+2*\Delta _fG_{CO_2}-\Delta _fG_{MnO_2}-2*\Delta _fG_{CO}\\\\\Delta G=0+2*-394.4-465.37-2*-137.3\\\\\Delta G=-979.57kJ

c)

\Delta G=\Delta _fG_{NH_3}+\Delta _fG_{HCl}-\Delta _fG_{NH_4Cl}\\\\\Delta G=16.7-95.3-(-342.81)\\\\\Delta G=264.21kJ

Regards.

6 0
3 years ago
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