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uranmaximum [27]

2 years ago

12

How many atoms of phosphorus are in 9.20 mol of copper(II) phosphate?

1 answer:

Anika [276]2 years ago

4 0

The number of atoms of phosphorus in copper (II) phosphate [Cu3 (PO4)2] is determined by multiplying first 9.20 by 2 since there are two phosphorus atoms in the compound and multiplying again with Avogadro's number. The answer is 1.108 x 10^25 atoms.

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How important is economic in your everyday life

docker41 [41]

Answer:

it is mega important.

Explanation:

3 0

2 years ago

Calculate the pH of the following:

Brums [2.3K]

The pH can be defined as the negative logarithm of the hydrogen ion concentration of the solution.

<h3>What is the pH?</h3>

What we call the pH can be defined as the negative logarithm of the hydrogen ion concentration of the solution. We are aware that we can use the relation [H+] [OH-] = 1 * 10^-14 to handle the enormity of this problem.

Now, let us go about solving the problems;

1. pH = -log(1 x 10-7) = 7

2. [H+]= 1 * 10^-14/ 1 x 10^-3

pH = -log( 1 * 10^-11)

pH = 11

3. pH = -log( 1 x 10^-2)

pH = 2

4. pH = -log( 1 x 10^-10)

pH = 10

5. [H+]= 1 * 10^-14/ 1 x 10^-8

[H+]= 1 * 10^-6

pH = 6

Learn more about pH:brainly.com/question/1528974

#SPJ1

6 0

1 year ago

If the average atomic mass of Rhodium is 102.91 amu, which isotope do you think has the greater percent abundance?

Illusion [34]

round up given mass to a whole number

$\\ \sf\longmapsto 102.91$

$\\ \sf\longmapsto 102.9$

$\\ \sf\longmapsto 103$

Hence our answer is Rhodium-103

8 0

2 years ago

he equilibrium constant, Kc , for the following reaction is 7.00×10-5 at 673 K.NH4I(s) NH3(g) + HI(g)If an equilibrium mixture o

frosja888 [35]

<u>Answer:</u> The number of moles of HI in the solution is $1.24\times 10{-3}$ moles.

<u>Explanation:</u>

We are given:

$K_c=7.00\times 10^{-5}\\n_{NH_3}=0.405mol\\n_{NH_4I}=1.45mol\\V=4.90L$

To calculate the concentration of a substance, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ ......(1)

Concentration of ammonia:

$[NH_3]=\frac{0.405mol}{4.90L}=0.083mol/L$

Concentration of ammonium iodide:

$[NH_4I]=\frac{1.45mol}{4.90L}=0.30mol/L$

For the given chemical reaction:

$NH_4I(s)\rightleftharpoons NH_3(g)+HI(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI][NH_3]}{[NH_4I]}$

Putting values in above equation, we get:

$7.0\times 10^{-5}=\frac{[HI]\times 0.083}{0.30}$

$[HI]=2.53\times 10^{-4}$

Calculating the moles of hydrogen iodide by using equation 1, we get:

$2.53\times 10^{-5}=\frac{\text{Moles of HI}}{4.9}\\\\\text{Moles of HI}=1.24\times 10^{-3}$

Hence, the number of moles of HI in the solution is $1.24\times 10{-3}$ moles.

3 0

2 years ago

Given 2.0 moles of nitrogen and plenty of hydrogen, how many moles of NH3 are formed?

kolezko [41]

3.0 moles

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3 0

2 years ago