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4 years ago

what is the oxidation number for chromium in each of these compounds. show your work. A. K2CrO4 B.Cr2O3

Chemistry

1 answer:

Dmitry [639]4 years ago

7 0

Answer:

A. + 6.

B. + 3.

Explanation:

A. K₂CrO₄:

To calculate the oxidation no. of an element in a compound:

We have that the sum of the oxidation no. of different elements in the compound multiplied by its no. is equal to the overall charge of the compound.

∵ 2(oxidation no. of K) + (oxidation no. of Cr) + 4(oxidation no. of O) = 0.

oxidation no. of K = + 1, oxidation no. of O = - 2.

∴ 2(+ 1) + (oxidation no. of Cr) +4(- 2) = 0.

∴ 2 + (oxidation no. of Cr) - 8 = 0.

∴ (oxidation no. of Cr) - 6 = 0.

∴ oxidation no. of Cr = + 6.

B. Cr₂O₃:

∵ 2(oxidation no. of Cr) + 3(oxidation no. of O) = 0.

oxidation no. of O = - 2.

∴ 2(oxidation no. of Cr) + 3(- 2) = 0.

∴ 2(oxidation no. of Cr) - 6 = 0.

∴ 2(oxidation no. of Cr) = + 6.

∴ (oxidation no. of Cr) = + 6/2 = + 3.

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What is he empirical formula for the compound that is of 1.85 moles of nitrogen and 4.63 miles of oxygen

BartSMP [9]

The empirical formula is N₂O₅.

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles, so our job is to calculate the molar ratio of N:O.

I like to summarize the calculations in a table.

Element Moles Ratio¹ ×2² Integers³

N 1.85 1 2 2

O 4.63 2.503 5.005 5

¹To get the molar ratio, you divide each number of moles by the smallest number (1.85).

²Multiply these values by a number (2) that makes the numbers in the ratio close to integers.

³Round off the number in the ratio to integers (2 and 5).

The empirical formula is N₂O₅.

4 0

3 years ago

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Vladimir79 [104]

The scientist can analyze that data and help other scientists with his research

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4 years ago

Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction: 2NO2 → 2NO + O2 In a particular experiment at 300 °C, [

Setler [38]

Answer:

$rate=-1.75x10^{-5}\frac{M}{s}$

Explanation:

Hello,

In this case, for the given information, we can compute the rate of disappearance of NO₂ by using the following rate relationship:

$rate=\frac{1}{2}*\frac{C_f-C_0}{t_f-t_0}$

Whereas it is multiplied by the the inverse of the stoichiometric coefficient of NO₂ in the reaction that is 2. Moreover, the subscript f is referred to the final condition and the subscript 0 to the initial condition, thus, we obtain:

$rate=\frac{1}{2}*\frac{0.00650M-0.0100M}{100s-0s}\\\\rate=-1.75x10^{-5}\frac{M}{s}$