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3 years ago

What is the mass in grams of 9.76 × 1012 atoms of naturally occurring beryllium?

Chemistry

2 answers:

Ksivusya [100]3 years ago

8 0

The atomic mass of beryllium (Be) is 9 g/mole

Now, 1 mole of any substance contains avogadro's number of that substance. Therefore:

1 mole of Be contains 6.023 10^23 atoms of Be

In other words:

9 grams of Be contains 6.023*10^23 atoms of Be

Therefore, the mass corresponding to 9.76 * 10^12 atoms of Be is:

= 9 g * 9.76 10^12 atoms/6.023*10^23 atoms

= 1.458 * 10^-10 g (or) 1.46 * 10^-10 g

Reika [66]3 years ago

6 0

Answer: The mass of given number of atoms of beryllium is $1.46\times 10^{-10}g$

Explanation:

Beryllium is the 4th element of the periodic table.

We know that:

Molar mass of beryllium = 9 g/mol

We are given:

Number of beryllium atoms = $9.76\times 10^{12}$

According to mole concept:

$6.022\times 10^{23}$ number of molecules occupy 1 mole of a gas.

As, $6.022\times 10^{23}$ number of beryllium atoms has a mass of 9 grams.

So, $9.76\times 10^{12}$ atoms of beryllium will have a mass of = $\frac{9g}{6.022\times 10^{23}}\times 9.76\times 10^{12}=1.46\times 10^{-10}g$

Hence, the mass of given number of atoms of beryllium is $1.46\times 10^{-10}g$

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3 years ago

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of

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NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa

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Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L

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pH using Henderson - Hasselbalch equation:

pH = pKa + log ([salt]/[acid])

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3 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

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