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blondinia [14]
3 years ago
15

How many moles of silver are equivalent to 2.408 × 1024 atoms? NEED AN ANSWER A.S.A.P

Chemistry
1 answer:
ExtremeBDS [4]3 years ago
6 0

1 mol of silver means 6.02 x 10²³ Ag atoms

number of moles of silver

= 2.408 × 10²⁴ Ag atoms  x   <u>        1 mol  Ag              </u>

                                            6.02 x 10²³ Ag atoms

= 0.4 x 10

<u>= 4 moles</u> Ag


<u>4 moles of silver</u> are equivalent to 2.408 × 10²⁴ atoms

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Brrunno [24]

Answer:

A)

<u>4, 7, 4, 6</u>

B)

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Explanation:

NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)

__↑______↑

8.00 mol | 14.00 mol

________________

NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)

You can turn this into a system of variables which are solvable.

To do this, create variables for the coefficients of each compound in the reaction respectively.

a(NH_{3}(g)) + b(O_{2}(g)) → \\c(NO_{2}) + d(H_{2}O(g))

Because to be balanced, the count of atoms in each element of the compound correspond to the coefficient of the variable in that compound so that the count of the left (reactant) side is set equal to the right (product) side.

a corresponds to the coefficient of the first compound, b corresponds to the coefficient of the second compound, c corresponds to the coefficient of the third compound, and d corresponds to the coefficient of the fourth compound.

(Reactant = Product)

Reactant: 1a [N] Product: 1c.

Reactant: 3a [H] Product: 2d.

Reactant: 2b [O] Product: 2c + 1d.

Thus the system is:

1a = 1c

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Then just use the substitution methods to solve.

3 0
2 years ago
Calculate the missing variables in each experiment below using Avogadro’s law.
blagie [28]

Answer:

The answer to your question is: letter c

Explanation:

Data

V1 = 612 ml    n1 = 9.11 mol

V2 = 123 ml    n2 = ?

Formula

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                                         n2 = \frac{n1V2}{V1}

                                         n2 = \frac{(9.11)((123)}{(612)}

                                                n2 = 1.83 mol                                                

5 0
3 years ago
There are only two isotopes of gallium: ^69Ga (68.926 amu) and ^71Ga(70.025 amu). The average atomic mass of gallium is 69.723 a
SVEN [57.7K]
The question ask for the percentage of the abundance of galium-69 where there is two isotopes of galium: the 69Ga and the 71Ga. The average atomic mass of gallium is 69.723 amu. So the formula would be <span>69.723amu=(%x)∗(68.926amu)+(1−%x)∗(70.025amu) and the answer to this is 1.58%</span>
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3 years ago
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3 years ago
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