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creativ13 [48]
1 year ago
14

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

report your answer with the correct number of significant figures
Chemistry
1 answer:
Gnom [1K]1 year ago
7 0

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

#SPJ4

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Answer:

hey are the final link in the energy flow in a food chain or a food web. They are fungi and animals that feed on dead organic mattet.

Explanaticuz im naruto uzamaki

5 0
3 years ago
Locations Cost of land Cost of equipment Cost of mining and reclamation Time taken to mine the area 1 $100,000 $10,000 $5,000 pe
mario62 [17]

Answer:

Location 2 will cost the least to the company to mine.

Explanation:

a) Data and Calculations:

Locations  Cost of land     Cost of       Cost of mining      Time taken to

                                        equipment   and reclamation    mine the area

1                 $100,000      $10,000         $5,000 per day        30 days

2                 $35,000       $6,000          $4,500 per day        45 days

3                 $30,000       $7,500          $3,500 per day       120 days

4                 $40,500       $8,000          $7,000 per day        65 days

Locations  Cost of land     Cost of       Cost of mining                      Total

                                        equipment   and reclamation                    Costs

1                 $100,000      $10,000         $150,000 ($5,000 * 30)  $260,000

2                 $35,000       $6,000         $202,500 ($4,500 * 45)   $243,500

3                 $30,000       $7,500         $420,000 ($3,500 * 120)  $457,500

4                 $40,500       $8,000         $445,000 ($7,000 * 65)   $503,500

7 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
Help pleaseeee. over dilutions in chemistry
atroni [7]

Answer:

5 L

Explanation:

We'll begin by calculating the molarity of the CaCl₂ solution. This can be obtained as follow:

Mole of CaCl₂ = 0.5 mole

Volume = 2 L

Molarity =?

Molarity = mole /Volume

Molarity = 0.5 / 2

Molarity = 0.25 M

Finally, we shall determine the volume of the diluted solution. This can be obtained as follow:

Molarity of stock solution (M₁) = 0.25 M

Volume of stock solution (V₁) = 2 L

Molarity of diluted solution (M₂) = 0.1 M

Volume of diluted solution (V₂) =?

M₁V₁ = M₂V₂

0.25 × 2 = 0.1 × V₂

0.5 = 0.1 × V₂

Divide both side by 0.1

V₂ = 0.5 / 0.1

V₂ = 5 L

Thus the volume of the diluted solution is 5 L

5 0
3 years ago
Which of the following is an example of physical change
BartSMP [9]
A physical change is like freezing water and it turns into Ice.
8 0
3 years ago
Read 2 more answers
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