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Allushta [10]
3 years ago
12

One brand of vinegar has a pH of 4.5. Another brand has a pH of 5.0. The equation for the pH of a substance is pH = –log[H+], wh

ere H+ is the concentration of hydrogen ions. What is the approximate difference in the concentration of hydrogen ions between the two brands of vinegar?
A. 2.2 x10^-5
B. 3.2x10^-1
C. 3.2x10^1
D.6.8x10^4
Chemistry
2 answers:
grigory [225]3 years ago
7 0
[H+] in first brand:
4.5 = -log([H+])
[H+] = 10^(-4.5)

[H+] in second brand:
5 = -log[H+]
[H+] = 10^(-5)

Difference = 10^(-4.5) - 10^(-5)
= 2.2 x 10⁻⁵

The answer is A.
LuckyWell [14K]3 years ago
6 0

Answer:

A. 2.2 x10^-5

Explanation:

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A solution has a poh of 7. 1 at 10∘c. what is the ph of the solution given that kw=2. 93×10−15 at this temperature? remember to
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A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .

It is given that,

pOH of solution = 7.1

Kw =2.93×10^(-15)

Firstly, we will calculate the value of pKw

The expression which we used to calculate the pKw is,

pKw=-log [Kw]

Now by putting the value of Kw in this expression,

pKw =−log{2.93×10^(-15)}

pKw =15log(2.93)

pKw=14.5

Now we have to calculate the pH of the solution.

As we know that,

pH+pOH=pKw

Now put all the given values in this formula,

pH+7.1=14.5

pH=7.4

Therefore, we find the value of pH of the solution is, 7.4.

learn more about pH value:

brainly.com/question/12942138

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7 0
1 year ago
What mass of butane in grams is necessary to produce 1.5×103 kJ1.5×103 kJ of heat? What mass of CO2CO2 is produced? Assume the r
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32.8 g of Butane is required and 99.3 g of CO₂ is produced

<u>Explanation:</u>

The above mentioned reaction can be written as,

C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g)     where ΔH (rxn)= -2658 kJ

It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.

That is

$\frac{1500}{2658}=0.564 \text { moles }    of butane reacted

Now this moles is converted into mass by multiplying it with its molar mass  = 0.564 mol × 58.122 g / mol

                     = 32.8 g of butane.

Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol

                                        = 99.3 g of CO₂

Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced

7 0
3 years ago
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