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nirvana33 [79]
3 years ago
13

How many moles of gold are equivalent to 1.204 × 1024 atoms?

Chemistry
2 answers:
Zinaida [17]3 years ago
8 0

How many moles of gold are equivalent to 1.204 × 1024 atoms?

0.2

0.5

2 CORRECT ANSWER

5

C) 2 Is the correct answer, I took the test and it was correct.

LUCKY_DIMON [66]3 years ago
7 0

Explanation:

It is known that the value of Avogadro's number is 6.022 \times 10^{23}.

Therefore, calculate moles of gold atoms as follows.

    Moles of gold atom = \frac{given number of atoms}{Avogadro's number}

                                    = \frac{1.204 \times 10^{24}}{6.022 \times 10^{23}mol^{-1}}

                                    = 0.199 \times 10 mol

                                    = 1.99 mol

Thus, we can conclude that 1.99 mol of gold atoms are equivalent to 1.204 \times 10^{24} atoms.

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in this series, Gold, Silver and Mercury are below Copper, this is, they are less reactive than Cu. this means none of them can displace it from its compounds.

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1) How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mm Hg?
RSB [31]

Answer : The number of molecules present in nitrogen gas are, 3.48\times 10^{13}

Explanation :

First we have to calculate the moles of nitrogen gas by using ideal gas equation.

PV=nRT

where,

P = Pressure of N_2 gas = 1.00\times 10^{-6}mmHg=1.32\times 10^{-9}atm      (1 atm = 760 mmHg)

V = Volume of N_2 gas = 985 mL = 0.982 L    (1 L = 1000 mL)

n = number of moles N_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 0.0^oC=273+0.0=273K

Now put all the given values in above equation, we get:

(1.32\times 10^{-9}atm)\times 0.982L=n\times (0.0821L.atm/mol.K)\times 273K

n=5.78\times 10^{-11}mol

Now we have to calculate the number of molecules present in nitrogen gas.

As we know that 1 mole of substance contains 6.022\times 10^{23} number of molecules.

As, 1 mole of N_2 gas contains 6.022\times 10^{23} number of molecules

So, 5.78\times 10^{-11} mole of N_2 gas contains (5.78\times 10^{-11})\times (6.022\times 10^{23})=3.48\times 10^{13} number of molecules

Therefore, the number of molecules present in nitrogen gas are, 3.48\times 10^{13}

8 0
3 years ago
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