First, we have to remember the molarity formula:
Part 1:
In this case, our solute is sodium nitrate (NaNO3), and we have the mass dissolved in water, then we have to convert grams to moles. For that, we need the molecular weight:
Then, we calculate the moles present in the solution:
Now, we have the necessary data to calculate the molarity (with the solution volume of 200 mL):
The molarity of this solution equals 0.2339 M.
Part 2:
In this case, we have the same amount (in moles and mass) of sodium nitrate, but a different volume of solution, then we only have to change it:
So, the molarity of this solution is 0.1701 M.
Answer:
m H2(g) = 2.241 g H2(g)
Explanation:
- 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)
limit reagent:
∴ Mw Al = 26.982 g/mol
∴ Mw H2SO4 = 98.0785 g/mol
⇒ n Al = (20 g Al)×(mol/26.982 g) = 0.7412 mol Al
⇒ n H2SO4 = ( 115 g H2SO4 )×(mol/98.0785 g) = 1.173 mol H2SO4
⇒ n H2 = (0.7412 mol Al)×(3 mol H2/ 2 mol Al) = 1.112 mol H2
∴ Mw H2 = 2.016 g/mol
⇒ g H2 = (1.112 mol H2)×(2.016 g/mol) = 2.241 g H2
Your answer is 1-3-2-4 , grass sand water rock
When Iodine-131 emits a β particle will produce Xe-131
<h3>Further explanation
</h3>
Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,
- alpha α particles ₂He⁴
- beta β ₋₁e⁰ particles
- gamma particles ₀γ⁰
- positron particles ₁e⁰
- neutron ₀n¹
So for reaction Iodine-131 :
