When equilibrium has been reached so, according to this formula we can get the specific heat of the unknown metal and from it, we can define the metal as each metal has its specific heat:

Mw*Cw*ΔTw = Mm*Cm*ΔTm

when
Mw → mass of water
Cw → specific heat of water
ΔTw → difference in temperature for water

Mm→ mass of metal
Cw→ specific heat of the metal
ΔTm → difference in temperature for metal

by substitution:

100g * 4.18 * (40-39.8) = 8.23 g * Cm * (50-40)

∴ Cm = 83.6 / 82.3 = 1.02 J/g.°C

when the Cm of the Magnesium ∴ the unknown metal is Mg

6 0

3 years ago

How does increasing the temperature affect a reaction?

Luden [163]

Answer:A. An increase in temperature increases the reaction rate.

Explanation:because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction.

4 0

3 years ago

How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut

san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = 8 = Concentration in Ethyl acetate / Concentration in water

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

Where X is the amount of material in grams that comes to the organic phase.

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

In the second extraction, you will extract 0.50g of material

Thus, after the two extraction you will lost:

4.44g + 0.50g = 4.94g of material

6 0

3 years ago