Answer:
709 g
Step-by-step explanation:
a) Balanced equation
Normally, we would need a balanced chemical equation.
However, we can get by with a partial equation, as log as carbon atoms are balanced.
We know we will need an equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.
M_r: 30.07 236.74
C₂H₆ + … ⟶ C₂Cl₆ + …
m/g: 90.0
(i) Calculate the moles of C₂H₆
n = 90.0 g C₂H₆ × (1 mol C₂H₆ /30.07 g C₂H₆)
= 2.993 mol C₂H₆
(ii) Calculate the moles of C₂Cl₆
The molar ratio is (1 mol C₂Cl₆/1 mol C₂H₆)
n = 2.993 mol C₂H₆ × (1 mol C₂Cl₆/1 mol C₂H₆)
= 2.993 mol C₂Cl₆
(iii) Calculate the mass of C₂Cl₆
m = 2.993 mol C₂Cl₆ × (236.74 g C₂Cl₆/1 mol C₂Cl₆)
m = 709 g C₂Cl₆
The reaction produces 709 g C₂Cl₆.
Potential energy or kinetic energy.
<span>Here we are asked to know the type of bond
between a glycosidic bond. A glycosidic bond is a type of bond that
exists between a carbohydrate molecule to another carbohydrate molecule. A
glycosidic bond creates between two monosaccharides can also be called as an
ether bond.</span>
Answer:
E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g)
Explanation:
Which ONE of the following is an oxidation–reduction reaction?
A) PbCO₃(s) + 2 HNO₃(aq) ⇒ Pb(NO₃)₂(aq) + CO₂(g) + H₂O(l). NO. All the elements keep the same oxidation numbers.
B) Na₂O(s) + H₂O(l) ⇒ 2 NaOH(aq). NO. All the elements keep the same oxidation numbers.
C) SO₃(g) + H₂O(l) ⇒ H₂SO₄(aq). NO. All the elements keep the same oxidation numbers.
D) CO₂(g) + H₂O(l) ⇒ H₂CO₃(aq). NO. All the elements keep the same oxidation numbers.
E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g). YES. <u>C is reduced</u> and <u>H is oxidized</u>.
