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3 years ago

An atom of gold has a mass of 3.271 x 10^-22g. how many atoms of gold are in 5.00g of gold?

Chemistry

2 answers:

Lostsunrise [7]3 years ago

5 0

Answer:

number of gold atoms = $1.528X10^{22}$

Explanation:

The mass of one atom of gold is given to be = $3.271X10^{-22}$

The mass of gold is given to be = $5.00g$

In order to determine the number of atoms present in the 5 g of gold, we will divide the mass of the gold with the mass of each atom of the gold.

the number of atoms = $\frac{massofgold}{massofatom}=\frac{5}{3.271X10^{-22}} =1.528X10^{22}$

Thus number of gold atoms = $1.528X10^{22}$

ra1l [238]3 years ago

4 0

5.00g Au x(1 atom Au/3.271x10^-22 g) = 1.52858x10^22 atoms
Since one of your given terms has only 3 significant figures (5.00g), your answer should only have 3 sig figs showing: 1.53x10^22 atoms Au

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Part 1. A chemist reacted 12.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat

galben [10]

The mass of NaCl needed for the reaction is 91.61 g

We'll begin by calculating the number of mole of F₂ that reacted.

Volume (V) = 12 L

Temperature (T) = 280 K

Pressure (P) = 1.5 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =?

PV = nRT

1.5 × 12 = n × 0.0821 × 280

18 = n × 22.988

Divide both side by 22.988

n = 18 / 22.988

n = 0.783 mole

Next, we shall determine the mole of NaCl needed for the reaction.

F₂ + 2NaCl —> Cl₂ + 2NaF

From the balanced equation above,

1 mole of F₂ reacted with 2 moles of NaCl.

Therefore,

0.783 mole F₂ will react with = 0.783 × 2 = 1.566 moles of NaCl.

Finally, we shall determine the mass of 1.566 moles of NaCl.

Mole = 1.566 moles

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =?

Mass = mole × molar mass

Mass of NaCl = 1.566 × 58.5

Mass of NaCl = 91.61 g

Therefore, the mass of NaCl needed for the reaction is 91.61 g

Learn more about stiochoimetry: brainly.com/question/25830314

6 0

2 years ago

At what temperature would 2.10moles of N2 gas have a pressure of 1.25atm and fill a 25.0 L tank

hodyreva [135]

Answer:

$\large \boxed{\text{-92 $^{\circ}$C}}$

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

$\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}$

2. Temperature in degrees Celsius

$\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}$