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kherson [118]
3 years ago
10

What is the [OH-] of a solution with a pH of 1.4?

Chemistry
1 answer:
denis23 [38]3 years ago
3 0

Answer:

If a solution has pH=8 , then its pOH is 14−8=6 and the corresponding concentration of hydroxyl OH− ions is 10−6 mol per litre.

Explanation:

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Which of the following is a true statement about the water cycle?
Serggg [28]
I think its C: due to mixing and oceanic.....
3 0
3 years ago
If 3.0g of H20 actually forms what is the percentage yield
Sati [7]
Percent yield is expressed as the ratio of the actual yield and the theoretical yield of the reaction multiplied by 100 to get the percent value. The actual yield is usually given in a problem. The theoretical yield is calculated from the reaction. For this problem, it cannot be solved since we cannot obtain the theoretical yield.
5 0
3 years ago
A student uses visible spectrophotometry to determine the concentration of CoCl2(aq) in a sample solution. First the student pre
gladu [14]

Answer:

A

Explanation:

2.60 g of CoCl2 was dissolved in water to make 100 mL of solution.

Number of moles of CoCl2 dissolved = mass/molar mass

                                = 2.60/130

                                   = 0.02 mole

Molar concentration of solutions = number of moles/volume (dm3)

                          = 0.02/0.1

                            = 0.200 M

Hence, the molar concentration of thesolution is 0.200 molar.

Correct option = A

3 0
3 years ago
What physical property is changed when a waitress pours coffee from a pot into four mugs?
Mandarinka [93]
The correct answer is volume
3 0
3 years ago
Constants: A: MW = 150 g/mol; B: = MW 100 g/mol; C: MW = 200 g/mol. 2.0 g C was made from 4.5 g A and 4.0 g
Neko [114]

Answer:

a. 100%

b. 133%

c. 300%

Explanation:

To find yield first we need to determine theoretical yield converting each reactant to moles and find limitng reactant for each reaction:

<em>Moles A:</em>

4.5g * (1mol / 150g) = 0.03 moles

<em>Moles B:</em>

4.0g * (1mol / 100g) = 0.04 moles

a. For a complete reaction of 0.03 moles of A are needed:

0.03 moles A * (1 mole B / 3 moles A) = 0.01 moles of B

As there are 0.04 moles of B, A is limiting reactant.

Theoretical moles and mass of C are:

0.03 moles A * (1 mole C / 3 moles A) = 0.01 moles of C.

0.01 moles of C * (200g / mol) = 2g are produced.

Yield is:

2g / 2g * 100 = 100%

b. For a complete reaction of 0.03 moles of A are needed:

0.03 moles A * (3 mole B / 2 moles A) = 0.045 moles of B

As there are 0.04 moles of B, B is limiting reactant.

Theoretical moles and mass of C are:

0.04 moles B * (1 mole C / 3 moles B) = 0.0133 moles of C.

0.0133 moles of C * (200g / mol) = 2.67g are produced.

Yield is:

2.67g / 2g * 100 = 133%

c. For a complete reaction of 0.03 moles of A are needed:

0.03 moles A * (1 mole B / 1 moles A) = 0.03 moles of B

As there are 0.04 moles of B, A is limiting reactant.

Theoretical moles and mass of C are:

0.03 moles A * (1 mole C / 1 moles A) = 0.03 moles of C.

0.03 moles of C * (200g / mol) = 6g are produced.

Yield is:

6g / 2g * 100 = 300%

4 0
2 years ago
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