Answer:
24e⁻ are transferred by the reaction of respiration.
Explanation:
C₆H₁₂O₆ + 6O₂ → 6 H₂O + 6CO₂
This is the reaction for the respiration process.
In this redox, oxygen acts with 0 in the oxidation state on the reactant side, and -2 in the product side - REDUCTION
Carbon acts with 0 in the glucose (cause it is neutral), on the reactant side and it has +4, on the product side - OXIDATION
6C → 6C⁴⁺ + 24e⁻
In reactant side we have a neutral carbon, so as in the product side we have a carbon with +4, it had to lose 4e⁻ to get oxidized, but we have 6 carbons, so finally carbon has lost 24 e⁻
6O⁻² + 6O₂ + 24e⁻ → 6O₂²⁻ + 6O⁻²
In reactant side, we have 6 oxygen from the glucose (oxidation state of -2) and the diatomic molecule, with no charge (ground state), so in the product side, we have the oxygen from the dioxide with -2 and the oxygen from the water, also with -2 at the oxidation state. Finally the global charge for the product side is -36, and in reactant side is -12, so it has to win 24 e⁻ (those that were released by the C) to be reduced.
Answer:
H₂SO₄
Explanation:
We have a compound formed by 0.475 g H, 7.557 g S, 15.107 g O. In order to determine the empirical formula, we have to follow a series of steps.
Step 1: Calculate the total mass of the compound
Total mass = mass H + mass S + mass O = 0.475 g + 7.557 g + 15.107 g
Total mass = 23.139 g
Step 2: Determine the percent composition.
H: (0.475g/23.139g) × 100% = 2.05%
S: (7.557g/23.139g) × 100% = 32.66%
O: (15.107g/23.139g) × 100% = 65.29%
Step 3: Divide each percentage by the atomic mass of the element
H: 2.05/1.01 = 2.03
S: 32.66/32.07 = 1.018
O: 65.29/16.00 = 4.081
Step 4: Divide all the numbers by the smallest one
H: 2.03/1.018 ≈ 2
S: 1.018/1.018 = 1
O: 4.081/1.018 ≈ 4
The empirical formula of the compound is H₂SO₄.
The first element to have 4d electrons in its electron configuration is Yttrium with an atomic number of 39 and the chemical symbol Y. The electron configuration of an atom refers to the distribution of its electrons in the atomic orbitals. These orbitals are represented by letters s,p,d,f and so on. The electron configuration of Yttrium is [Kr] 4d¹ 5s², wherein [Kr] refers to the electron configuration of the noble gas, Krypton.
Because they frequently have a long half-lives, therefore his stay in the middle is long.
Number 4
If you notice any mistake in my english, please let me know, because i am not native.
Answer:
The partial pressure of neon in the vessel was 239 torr.
Explanation:
In all cases involving gas mixtures, the total gas pressure is related to the partial pressures, that is, the pressures of the individual gaseous components of the mixture. Put simply, the partial pressure of a gas is the pressure it exerts on a mixture of gases.
Dalton's law states that the total pressure of a mixture of gases is equal to the sum of the pressures that each gas would exert if it were alone. Then:
PT= P1 + P2 + P3 + P4…+ Pn
where n is the amount of gases present in the mixture.
In this case:
PT=PN₂ + PAr + PHe + PNe
where:
- PT= 987 torr
- PN₂= 44 torr
- PAr= 486 torr
- PHe= 218 torr
- PNe= ?
Replacing:
987 torr= 44 torr + 486 torr + 218 torr + PNe
Solving:
987 torr= 748 torr + PNe
PNe= 987 torr - 748 torr
PNe= 239 torr
<u><em>The partial pressure of neon in the vessel was 239 torr.</em></u>
