Answer:
Hi! I believe this is your answer:
52 kilograms
Hope this helps, sorry if it's wrong!
Explanation:
Answer:
The arrow points from the reactants to the products, so just follow the arrows.
Explanation:
some have the reactants on the left and the products on the right, and others are the opposite... just know that
reactants---------> products
or
products<-----------reactants
Answer:
Initial volume of the container (V1) = 1.27 L (Approx)
Explanation:
Given:
Number of mol (n1) = 5.67 x 10⁻²
Number of mol (n2) = (5.67 +2.95) x 10⁻² = 8.62 x 10⁻²
New volume (V2) = 1.93 L
Find:
Initial volume of the container (V1)
Computation:
Using Avogadro's law
V1 / n1 = V2 / n2
V1 / 5.67 x 10⁻² = 1.93 / 8.62 x 10⁻²
V1 = 10.9431 / 8.62
Initial volume of the container (V1) = 1.2695
Initial volume of the container (V1) = 1.27 L (Approx)
Answer:
89.04 g of NaNO₃.
Explanation:
We'll begin by converting 838 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
838 mL = 838 mL × 1 L / 1000 mL
838 mL = 0.838 L
Next, we shall determine the number of mole of NaNO₃ in the solution. This can be obtained as follow:
Volume = 0.838 L
Molarity = 1.25 M
Mole of NaNO₃ =?
Mole = Molarity × volume
Mole of NaNO₃ = 1.25 × 0.838
Mole of NaNO₃ = 1.0475 mole
Finally, we shall determine the mass of NaNO₃ needed to prepare the solution. This can be obtained as follow:
Mole of NaNO₃ = 1.0475 mole
Molar mass of NaNO₃ = 23 + 14 + (16×3)
= 23 + 14 + 48
= 85 g/mol
Mass of NaNO₃ =?
Mass = mole × molar mass
Mass of NaNO₃ = 1.0475 × 85
Mass of NaNO₃ = 89.04 g
Therefore, 89.04 g of NaNO₃ is needed to prepare the solution.
