How would you prepare 500 cm3 of 2.0 M solution of Na2CO3 solution from its anhydrous salt? [C = 12, O = 16, Na = 23

The answer to your question is given below

Explanation:

We'll begin by converting 500 cm³ to L. This can be obtained as follow:

1000 cm³ = 1 L

Therefore,

500 cm³ = 500 cm³× 1 L / 1000 cm³

500 cm³ = 0.5 L

Next, we shall determine the number of mole of Na₂CO₃ in the solution. This can be obtained as follow:

Volume = 0.5 L

Molarity = 2 M

Mole of Na₂CO₃ =?

Molarity = mole /Volume

2 = Mole of Na₂CO₃ / 0.5

Cross multiply

Mole of Na₂CO₃ = 2 × 0.5

Mole of Na₂CO₃ = 1 mole

Finally, we shall determine the mass of 1 mole of Na₂CO₃. This can be obtained as follow:

Mole of Na₂CO₃ = 1 mole

Molar mass of Na₂CO₃ = (23×2) + 12 + (16×3)

= 46 + 12 + 48

= 106 g/mol

Mass of Na₂CO₃ =?

Mass = mole × molar mass

Mass of Na₂CO₃ = 1 × 106

Mass of Na₂CO₃ = 106 g

Thus, to prepare the solution, weigh 106 g of Na₂CO₃ and dissolve in 500 cm³ (i.e 0.5 L) of water.

the answer is either b or c

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.