Chemistry, 21.02.2020 16:20 briannagiddens

Give the oxidation number of the element indicated for each compound.

a) carbon in Na2C2O4
b) nitrogen in HNO3
c) sulfur in H2SO4
d) carbon in HCO3 −
e) manganese in KMnO4
f) chromium in Cr2O7 2−

Answers: 2

Show answers

Answers

Answer from: skrillex88

a. C: +3 ; b. N: +5 ; c. S:+6 ; d. C: +4; e. Mn: +7 ; f. Cr: +6.

Explanation:

Global charges in molecules is 0

You sum all the oxidation states to determine the oxidation state for the compound.

Na₂C₂O₄ → Sodium oxalate → Global charge: 0

Oxidation state for C: +3

HNO₃ → Nitric acid → Global charge: 0

Oxidation state for N: +5

H₂SO₄ → Sulfuric acid → Global charge: 0

Oxidation state for S: +6

HCO₃⁻ → Bicarbonate → Global charge: -1, this is an anion

Oxidation state for C: +4

KMnO₄ → Potassium permanganate → Global charge: 0

Oxidation state for Mn: +7

Cr₂O₇⁻ → Anion dichromate → Global charge: -2

Oxidation state for Cr: +6

Answer from: Quest

A. it generates energy from carbohydrates.b. it requires activation by oxygen.c. it is used by cells to dispose of excess electrons

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

Another question on Chemistry

Chemistry, 21.06.2019 22:30

Determine the wavelength of the light absorbed when an electron in a hydrogen atom makes a transition from an orbital in the n=3 level to an orbital in the n=7 level.

Answers: 2

Answer

Chemistry, 22.06.2019 00:30

You have 125g of a certain seasoning and are told that it contains 76.0 g of salt what is the percentage of salt by mass in this seasoning

Answers: 1

Answer

Chemistry, 22.06.2019 03:30

Predict the products butane (c4h10) + oxygen —>

Answers: 1

Answer

Chemistry, 22.06.2019 12:00

What is a possible quantum number set for an electron in the 3s orbital of a magnesium atom