Explanation:
A polymer forms because the hydrocarbon joins with itself in a polymerization reaction.
Answer:
58.6 % by mass of Na₂CO₃
Explanation:
This is the reaction:
Na₂CO₃ + MgCO₃ + 4HCl → MgCl₂ + 2NaCl + 2CO₂ + 2H₂O
Let's find out the moles of CO₂ produced, by the Ideal Gases Law
1.24 atm . 1.67 L = n . 0.082 . 299K
(1.24 atm . 1.67 L / 0.082 . 299K) = n
0.0844 moles = n
Ratio is 2:1, so 2 moles of dioxide were produced by 1 mol of sodium carbonate. Let's make a rule of three:
2 moles of CO₂ were produced by 1 mol of Na₂CO₃
Then, 0.0844 moles of Co₂ would beeen produced by (0.0844 .1)/2 = 0.0422 moles of Na₂CO₃.
Let's convert this moles into mass (mol . molar mass)
0.0422 mol . 106 g/mol = 4.47 g
Finally we can know the mass percent of sodium carbonate in the mixture
(Mass of compound /Total mass) . 100 → (4.47 g / 7.63g) . 100 = 58.6 %
When the amount of heat gained = the amount of heat loss
so, M*C*ΔTloses = M*C* ΔT gained
when here the water is gained heat as the Ti = 25°C and Tf= 28°C so it gains more heat.
∴( M * C * ΔT )W = (M*C*ΔT) Al
when Mw is the mass of water = 100 g
and C the specific heat capacity of water = 4.18
and ΔT the change in temperature for water= 28-25 = 3 ° C
and ΔT the change in temperature for Al = 100-28= 72°C
and M Al is the mass of Al block
C is the specific heat capacity of the block = 0.9
so by substitution:
100 g * 4.18*3 = M Al * 0.9*72
∴ the mass of Al block is = 100 g *4.18 / 0.9*72
= 19.35 g
Answer:
78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene
Explanation:
Moles of propene = 
According to reaction, 1 mole of propene gives 1 mole of propane.
Then 0.6538 moles of bromo-propane will give:
78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene.
