Answer:
3C3H8O2 and 16H2SO4
Explanation:
C3H8O2(aq) + K2Cr2O7(aq) → C3H4O4(aq) + Cr2(SO4)3(aq)
In acidic reaction you need to separate the half-reactions knows.
C3H8O2(aq) → C3H4O4(aq)
Cr2O7(2-) → Cr(+3) (aq)
Balance elements other than O and H. Add H2O to balance oxygen, balance hydrogen by adding protons (H+) and Balance the charge of each equation with electrons
C3H8O2(aq) + 2H2O(l) → C3H4O4(aq) + 8H(+) + 8e(-)
Cr2O7(2-) + 14H(+) + 6e(-) → 2Cr(+3) + 7H2O (aq)
Scale the reactions so that the electrons are equal. If not look for a common multiple. In this case is 24, 8*3=6*4. So for each equation multiply for the correct multiple. Then
3C3H8O2(aq) + 6H2O(l) → 3C3H4O4(aq) + 24H(+) + 24e(-)
4Cr2O7(2-) + 56H(+) + 24e(-) → 8Cr(+3) + 28H2O (aq)
Add the reactions and cancel out common terms and group the ions in compounds if u can.
3C3H8O2(aq) + 32H(+) + 4Cr2O7(2-) → 3C3H4O4(aq) + 4Cr2(SO4)3(aq) + 22H2O (aq)
We are in acidic medium so the H(+) must be from H2SO4. So 32H(+) is 16H2SO4 in reactives. Add SO4(2-) in products and balance
16H2SO4 +3C3H8O2(aq) + 4Cr2O7(2-) → 3C3H4O4(aq) 4Cr2(SO4)3(aq) + 22H2O (aq) + 4SO4(2-)
Then
16H2SO4 +3C3H8O2(aq) + 4Cr2O7(2-) → 3C3H4O4(aq) 4Cr2(SO4)3(aq) + 22H2O (aq) + + 2K2SO4 (aq)
The answer is B: friction.
A definition of friction is a resistance of an object against another, for example a tire against mud, the tire would add friction against the mud to pull away from it.
I don’t need brainliest but please just mark thanks to this answer that would help me a lot, have a nice day :)
Answer:
You can see in the above illustration of the periodic table that N is in Period 4, Group 5A, and p-block. Before reaching N, you will pass through the atomic orbitals 1s, 2s, and 2p. The electron configuration of N3- is N3-: 1s22s22p6 .
Explanation:
Answer:
Top left square is EE and bottom right square is Ee
