Answer: N = 2.78 × 10^23 atoms
There are N = 2.78 × 10^23 atoms in 70g of Au2cl6
Completed Question:
Calculate the number of gold atoms in a 70g sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and round it to significant digits
Explanation:
Given:
Molar mass of Au2cl6 = 303.33g/mol
Mass of Au2cl6 = 70g
Number of moles of Au2cl6 = 70g/303.33g/mol = 0.231mol
According to the chemical formula of Au2cl6,
1 mole of Au2cl6 contains 2 moles of Au
Number of moles of Au = 2 × 0.231mol = 0.462mole
There are 6.022 × 10^23 atoms in 1 mole of an element.
Number of Atom of gold in 0.462 mole of gold is:
N = 0.462 mol × 6.022 × 10^23 atoms/mol
N = 2.78 × 10^23 atoms
Answer:
sulfur-35
Explanation:
Sulfur-35 is a radioactive isotope that contains 19 neutrons.
Isotopes are represented with mass numbers. Mass number is the addition of number of proton and number of neutrons.
The number of proton in sulfur = 16
Number of neutron = 19
So, mass number = no. of protons + no. of neutrons
= 16 + 19
= 35
Hence, the correct answer is sulfur-35.
Answer: 287.8 cm3
Explanation:
Given that:
Initial volume of gas V1 = 350 cm3
Initial pressure of gas P1 = 740 mmHg
New volume V2 = ?
New pressure P2 = 900 mmHg
Since, pressure and volume are involved while temperature is constant, apply the formula for Boyle's law
P1V1 = P2V2
740 mmHg x 350 cm3 = 900mmHg x V2
V2 = (740 mmHg x 350 cm3) /900mmHg
V2 = 259000 mmHg cm3 / 900mmHg
V2 = 287.8 cm3
Thus, the gas will occupy 287.8 cubic centimeters at the new pressure.
