The answer is 4)change in enthalpy
For the reaction
3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH- (aq) → 3 Ni(s) + 2 CrO42-(aq) + 8 H2O(l) G = +87 kJ/mol
Given:
(Ni2+(aq) + 2 e- → Ni(s)) x 3 Ered = -0.28 V
(CrO42-(aq) + 4 H2O(l) + 3 e- → Cr(OH)3(s) + 5 OH-(aq)) x 2
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3 Ni2+(aq) + 6 e- → 3 Ni(s)
2 CrO42-(aq) + 8 H2O(l) + 6 e- → 2 Cr(OH)3(s) + 10 OH-(aq)
Answer: 0.72 litres of water is wasted in one day.
Explanation:
First you need to find out how many minutes are in a day. Do this by multiplying the number of minutes in an hour (60) by the number of hours in a day (24). 24 x 60 = 1440. If the faucet is dripping at 5 drops per minute, then multiply 5 by the number of minutes in a day (1440) to see how many drops drip in one day. 5 x 1440 = 7200. Now we need to figure out how many mL fo water that is. if 10 drops is 1 mL, then we need to divide the total number of drops (7200) by 10. 7200 divided by 10 is 720. That means 720 mL of water is dripping per day. Finally, we must convert mL to litres. There are 1000 mL in one litre, so divide 720 by 1000. The final answer is 0.72
Answer:
Butan-2-one
Explanation:
1. 1700 cm⁻¹
A strong peak near 1700 cm⁻¹ is almost certainly a carbonyl (C=O) group.
2. Triplet-quartet
A triplet-quartet pattern indicates an ethyl group.
The 2H quartet is a CH₂ adjacent to a CH₃. The peak normally occurs at δ 1.3, but it is shifted 1.2 ppm downfield to δ 2.47 by an adjacent C=O group.
The 3H triplet at δ 1.05 is the methyl group. It, too, is shifted downfield from its normal position at δ 0.9. The effect is smaller, because the methyl group is further from the carbonyl.
3. 3H(s) at δ 2.13
This indicates a CH₃ group with no adjacent hydrogen atoms.
It is shifted 0.8 ppm downfield to δ 2.13 by the adjacent C=O group.
4. Identification
The identified pieces are CH₃CH₂-, -(CO)-, and -CH₃. There is only one way to put them together: CH₃CH₂-(C=O)-CH₃.
The compound is butan-2-one.
Answer:
A
Explanation:
How many moles of HCI will just react with 0.424 g Ba(OH)2?
A) 4.94 x 10 mol
