Answer:
15.06 × 10²³ atoms of Li
Explanation:
Given data:
Number of moles of Li = 2.5 mol
Number of toms of Li = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
For 2.5 mol of Li:
1 mole of lithium = 6.022 × 10²³ atoms of Li
2.5 mol × 6.022 × 10²³ atoms of Li / 1 mol
15.06 × 10²³ atoms of Li
Answer:
electrolysis of brine
Explanation:
Rock salt deposits are usually mined; occasionally water is pumped down, and brine which contain 25 percent of sodium chloride is found
so d brine is electrolyzed to produce chlorine
Answer:
A compressed gas cylinder is filled with 5270 g of argon gas.
The pressure inside the cylinder is 2050 psi at a temperature of 18C.
The valve to the cylinder is opened and gas escapes until the pressure inside the cylinder is 650. psi and the temperature are 26 C.
How many grams of argon remains in the cylinder?
Explanation:
First, calculate the volume of argon gas that is present in the gas cylinder by using the ideal gas equation:
Mass of Ar gas is --- 5270g.
The number of moles of Ar gas:
Temperature T=(18+273)K=291K
Pressure P=2050psi
Volume V=?
Using this volume V=22.6L
Pressure=650psi=44.2atm
Temperature T= (26+273)K=299K
calculate number of moles "n" value:
Mass of 40.7mol of Ar gas:
Answer:
The mass of Ar gas becomes 1625.8g.
Answer:
1. The products of this reaction are ZnCl₂ and H₃PO₄.
2. 14.57 g.
Explanation:
<em>1. What would the products of this reaction be?</em>
- The balanced reaction between Zn₃(PO₄)₂ and HCl is represented as:
<em>Zn₃(PO₄)₂ + 6HCl → 3ZnCl₂ + 2H₃PO₄,</em>
It is clear that 1.0 mol of Zn₃(PO₄)₂ reacts with 6.0 mol of HCl to produce 3.0 mol of ZnCl₂ and 2.0 mol of H₃PO₄.
So, the products of this reaction are ZnCl₂ and H₃PO₄.
<em>2. If we produced 13.05 g of H₃PO₄, how many grams of hydrochloric acid would be need to start with?</em>
- Firstly, we should get the no. of moles (n) of 13.05 grams of H₃PO₄:
n = mass/molar mass = (13.05 g)/(97.994 g/mol) = 0.1332 mol.
<u><em>Using cross-multiplication:</em></u>
6.0 mol of HCl needed to produce → 2.0 mol of H₃PO₄, from stichiometry.
??? mol of HCl needed to produce → 0.1332 mol of H₃PO₄.
∴ The no. of moles of HCl needed = (6.0 mol)(0.1332 mol)/(2.0 mol) = 0.3995 mol.
∴ The mass of HCl needed = n*molar mass = (0.3995 mol)(36.46 g/mol) = 14.57 g.
<em>So, the grams of hydrochloric acid would be need to start with = 14.57 g.</em>
The term that refers to compounds that can form hydrates but do not contain water molecules is anhydrous.
