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>.
Non-volatile solutes such as salt raises the boiling point of water. Hope the answer helps! Good luck!
Answer:
To determine the amount of heat the gold has absorbed to melt, we simply multiply the mass of the block of ice to the heat of fusion of water which is given above. We calculate as follows:
Heat = 20.0 g (35.4 g)
Heat = 1290 J
2NaCN(s) + H₂SO₄(aq) --> Na₂SO₄(aq) + 2HCN(g)
The molar ratio between NaCN : HCN is 2:2 or 1:1
Mass of HCN = 16.7 g
Molar mass of HCN = 1 + 12 + 14 = 27 g/mol
Molar mass of NaCN = 49 g/mol
Therefore, the mass of NaCN is
16.7 g of HCN x 49 g/mol of NaCN / 27 g/mol of HCN = 30.3 grams of NaCN
Therefore, 30.3 grams of NaCN gives the lethal dose in the room.
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