It is A. Barium
Explanation: I did that already
0.115 M means that 0.115 moles of KBr are contained in a volume of 1000 ml, therefore a volume of 350 ml will have (0.115 × 0.35) = 04025 moles
From the formula of molarity moles = molarity × volume in liters
1 mole of KBr is equivalent to 119 g
Therefore, the mass = 0.04025 × 119 g = 4.79 g
Answer:
moles Ar in 20g = 0.500 mole Ar
Explanation:
moles = grams given / formula weight = 20g / 39.948g·mol⁻¹ = 0.500 mole Ar
Answer:
C = 18.29 g
Explanation:
Given data:
Mass of beryllium needed = ?
Mass of nitrogen = 18.9 g
Solution:
Chemical equation:
3Be + N₂ → Be₃N₂
now we will calculate the number of moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 18.9 g/ 28 g/mol
Number of moles = 0.675 mol
Now we will compare the moles of nitrogen and Be from balance chemical equation.
N₂ : Be
1 : 3
0.675 : 3/1×0.675 = 2.03 mol
Number of moles of Be needed are 2.03 mol.
Mass of Beryllium:
Mass = number of moles × molar mass
Mass = 2.03 mol × 9.01 g/mol
Mass = 18.29 g
