Answer:
15.14×10⁷ g
Explanation:
Given data:
Number of particles of NaCl = 1.56×10³⁰ particles
Mass of sodium chloride = ?
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.
1 mole = 6.022 × 10²³ particles
1.56×10³⁰ particles × 1 mol / 6.022 × 10²³ particles
0.259 ×10⁷ mol
Mass in gram:
Mass = number of moles × molar mass
Mass = 0.259 ×10⁷ mol × 58.44 g/mol
Mass = 15.14×10⁷ g
The answer is B. Your welcome.
The number 23.0702 has 6 significant figures.
To answer this, you must have the rules for significant figures,
1. All non-zero numbers ARE significant
2. Zeros between two non-zero digits ARE significant
3. Leading zeros are NOT significant
4. Trailing zeros are ONLY significant if the number has a decimal point
The number 23.0702 has 4 non-zero numbers(rule 1), and both zeros are between two non-zero digits(rule 2). Since all 6 figures in the number 23.0702 are significant, the number 23.0702 has 6 significant figures.
1. 12.992 L
2. 2.42 moles
3. 275.52 L
4. 567.844 g
<h3>Further explanation</h3>
Given
moles and volume at STP
Required
mass, volume and moles
Solution
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
1. 0.58 moles ammonia :
Volume = 0.58 moles x 22.4 L = 12.992 L
2. 77.5 grams of O₂ :
Moles = 77.5 grams x (1 mol/32 grams) = 2.42
3. 12.3 mole of Bromine gas :
Volume = 12.3 mole x (22.4 L/1 mole) = 275.52 L
4. 4.8 moles iron(II)chloride :
Mass = 4.48 moles x molar mass ( 126,751 g/mol) = 567.844 g
The equation is as follows;
2CH3OH(g) = 2CH4(g) + O2(g), ΔH= +252.8 kJ
From the equation; for the reaction to produce 2 moles of methane (32g) an energy of 252.8 kJ is released.
Therefore; for an energy of 82.3 kJ the number of moles that will be produced will be; = (2×82.3)/252.3
= 0.6524 moles
which is equivalent to 0.6524 × 16 = 10.438 g
There, the mass of CH4 produced will be 10.438 g
