Answer:
(A) is 0.0773 mol B2H6
(C) is 2.79 x 10^23 H atoms
Explanation:
Questions (A) and (B) are the same.
2.14 g B2H6 x (1 mol B2H6/27.668g B2H6) = 0.0773 mol B2H6 (A)
<u>27.668 is the molar mass of B2H6 calculated from the period table: </u>
(2 x 10.81) + (6 x 1.008) = 27.668
1.008 is the mass of H and 10.81 is the mass of B
(C)
0.0773 mol B2H6 x (6 mol H/ 1 mol B2H6) x (6.022 x 10^23 H atoms/1 mol H)
= 2.79 x 10^23 hydrogen atoms
Further Explanation:
- For every 1 mol of B2H6, there are 6 moles of H (indicated by the subscript)
- 6.022 x 10^23 is Avogrado's number and it equals to 1 mol of anything
- Avogrado's number can be in units of atoms, molecules, or particles
Moles of solute for both a and b are the same = 1 mol
<h3>Further explanation</h3>
Given
a 500 cm³ of solution, of concentration 2 mol/dm³
b 2 litres of solution, of concentration 0.5 mol/dm³
Required
moles of solute
Solution
Molarity shows the number of moles of solute in every 1 liter of solution or mmol in each ml of solution
Can be formulated :
a.
V = 500 cm³ = 0.5 L
M = 2 mol/L
n=moles = M x V
n = 2 mol/L x 0.5 L
n = 1 mol
b.
V = 2 L
M = 0.5 mol/L
n=moles = M x V
n = 0.5 mol/L x 2 L
n = 1 mol
Answer:
Molarity = 0.5 M
Explanation:
Given data:
Mass of NaCl = 2.7 g
Volume = 100 mL(100×10⁻³L)
Molarity of solution = ?
Solution:
Formula:
Molarity = number of moles / volume in litter
Number of moles:
Number of moles = mass/ molar mass
Number of moles = 2.7 g/ 58.5 g/mol
Number of moles = 0.05 mol
Molarity = number of moles / volume in litte
Molarity = 0.05 mol / 100×10⁻³L
Molarity = 0.5 M
C3H6 and CH4
The empirical number is found by looking at the ratio of the numbers. For example, H20 and H402 have the same empirical formula because the ratio remained similar. 2 H to 1 O.
