If you have 1.2754x1020 molecules of Barium nitrate (Ba(NO3)2), how many moles of barium nitrate do you have?

Chemistry

1 answer:

IgorLugansk [536]3 years ago

4 0

Answer:

0.00021186 moles of barium nitrate

Explanation:

The given parameter is as follows;

The number of molecules of Barium nitrate Ba(NO₃)₂ = 1.2754 × 10²⁰ molecules

The number of molecule present per mole of Barium nitrate, is given by Avogadro's number, 6.02 × 10²³

1 mole of a substance contains 6.02 × 10²³ molecules

The umber of moles in a substance = (The number of molecules of the substance)/(The number of molecules per mole)

Therefore the number of moles of barium nitrate in 1.2754 × 10²⁰ molecules of barium nitrate = (1.2754 × 10²⁰)/(6.02 × 10²³) ≈ 0.00021186 moles (rounding to 5 significant figures)

The number of moles of barium nitrate present in 1.2754 × 10²⁰ molecules of barium nitrate ≈ 0.00021186 moles

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18.A 2.50 L sample of dry air in a cylinder exerts a pressure of

Vlada [557]

Answer:

$\boxed{\text{7.50 L}}$

Explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

$p_{1}V_{1} = p_{2}V_{2}$

Data:

$\begin{array}{rclrcl}p_{1}& =& \text{3.00 atm}\qquad & V_{1} &= & \text{2.50 L} \\p_{2}& =& \text{1.00 atm}\qquad & V_{2} &= & ?\\\end{array}$

Calculations:

$\begin{array}{rcl}3.00 \times 2.50 & =& 1.00V_{2}\\7.500 & = & 1.00V_{2}\\V_{2} & = &\textbf{7.50 L}\\\end{array}\\\text{The volume of the gas is } \boxed{\textbf{7.50 L}}$