The mass of NiCl₂•6HO₂ needed to prepare a 0.035 M 500 mL solution of NiCl₂•6HO₂ is 4.165 g
<h3>What is molarity? </h3>
This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:
Molarity = mole / Volume
<h3>How to determine the mole of NiCl₂•6HO₂</h3>
- Molarity = 0.035 M
- Volume = 500 mL = 500 / 1000 = 0.5 L
Mole = Molarity × Volume
Mole of NiCl₂•6HO₂ = 0.035 × 0.5
Mole of NiCl₂•6HO₂ = 0.0175 mole
<h3>How to determine the mass of NiCl₂•6HO₂</h3>
- Mole of NiCl₂•6HO₂ = 0.0175 mole
- Molar mass of NiCl₂•6HO₂ = 238 g/mol
Mass = mole × molar mass
Mass of NiCl₂•6HO₂ = 0.0175 × 238
Mass of NiCl₂•6HO₂ = 4.165 g
Thus, 4.165 g of NiCl₂•6HO₂ is needed to prepare the solution
Learn more about molarity:
brainly.com/question/15370276
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The amount remaining at the end of 5 half-lives is 7.81×10¹³ g
From the question given above, the following data were obtained:
- Half-life (t½) = 5730 years
- Original amount (N₀) = 2.5×10¹⁵ g
- Number of half-lives (n) = 5
- Amount remaining (N) =?
The amount remaining can be obtained as follow:
N = 1/2ⁿ × N₀
N = 1/2⁵ × 2.5×10¹⁵
N = 1/32 × 2.5×10¹⁵
N = 0.03125 × 2.5×10¹⁵
N = 7.81×10¹³ g
Therefore, the amount remaining after 5 half-lives is 7.81×10¹³ g
Learn more about half-life: brainly.com/question/25783920
Solution :
Assuming all Al will be consumed in the reaction.
From the given equation, 2 mole of Al produce 2 mole of 
.
So, 1 mole of Al produce 1 mole of . .....statement 1)
Number of moles of Al in 258 gram of Al.
Therefore, from statement 1) number of moles of produced is also 9.556 moles.
