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3 years ago

What is the oxidation number of chromium in the ionic compound ammonium dichromate (nh4)2cr2o7?

1 answer:

4 0

The oxidation number of an element is the number of electrons that are gained or lost by the element to form a chemical bond.
the net oxidation number of the ion is its charge.
compound (NH₄)₂Cr₂O₇ is made of cation - NH₄⁺ and anion - Cr₂O₇²⁻
oxidation number of ion - -2
(oxidation number of Cr x 2 Cr atoms) +(oxidation number of O x 7 O atoms )= -2
oxidation number of Cr = y
oxidation number of O = -2
(y x 2) + (-2 x 7) = -2
2y - 14 = -2
2y = 12
y = + 6
therefore oxidation number of Cr is +6

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The mass of a single uranium atom is 4.70x10^-22 grams. How many uranium atoms would there be in 111 milligrams of uranium?

emmainna [20.7K]

2.4 x 10²² atoms

<h3>Further explanation</h3>

Atomic mass is the average atomic mass of all its isotopes

In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu

So the atomic mass obtained is the mass of the atom relative to the 12th carbon atom

mass single Uranium atom=4.7 x 10⁻²² g

then for 111 mg=0.111 g

$\tt \dfrac{0.111}{4.7\times 10^{-22}}=2.4\times 10^{22}$

5 0

2 years ago

A mixture of hydrogen and argon gases is maintained in a 6.47 L flask at a pressure of 3.43 atm and a temperature of 85 °C. If t

11Alexandr11 [23.1K]

Answer:

The mixture contains 8.23 g of Ar

Explanation:

Let's solve this with the Ideal Gases Law

Total pressure of a mixture = (Total moles . R . T) / V

We convert T° from °C to K → 85°C + 273 = 358K

3.43 atm = (Moles . 0.082 L.atm/mol.K . 358K) / 6.47L

(3.43 atm . 6.47L) / (0.082 L.atm/mol.K . 358K) = Moles

0.756= Total moles from the mixture

Moles of Ar + Moles of H₂ = 0.756 moles

Moles of Ar + 1.10 g / 2g/mol = 0.756 moles

Moles of Ar = 0.756 moles - 0.55 moles H₂ → 0.206

We convert the moles to g → 0.206 mol . 39.95 g / 1 mol = 8.23 g

8 0

2 years ago

Read 2 more answers

How does a nuclear plant operate

slava [35]

A process called fission

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3 years ago

Read 2 more answers

The decomposition of N2O4 into NO2 has Kp = 2. Some N2O4 is placed into an empty container, and the partial pressure of NO2 at e

Sunny_sXe [5.5K]

Answer: Option (B) is the correct answer.

Explanation:

Expression for the given decomposition reaction is as follows.

$N_{2}O_{4} \rightarrow 2NO_{2}$

Let us assume that x concentration of $N_{2}O_{4}$ is present at the initial stage. Therefore, according to the ICE table,

$N_{2}O_{4} \rightarrow 2NO_{2}$

Initial : x 0

Change : - 0.1 $2 \times 0.1$

Equilibrium : (x - 0.1) 0.2

Now, expression for $K_{p}$ of this reaction is as follows.

$K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}$

Putting the given values into the above formula as follows.

$K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}$

$2 = \frac{(0.2)^{2}}{(x - 0.1)}$

$2 \times (x - 0.1) = (0.2)^{2}$

x = 0.12

This means that $P_{N_{2}O_{4}}$ = x = 0.12 atm.

Thus, we can conclude that the initial pressure in the container prior to decomposition is 0.12 atm.

6 0

3 years ago

What factors do not affect nuclear reactions, but do affect chemical reaction

Fudgin [204]

Unlike nuclear reactions, nuclear reactions are not affected by changes in temperature,
pressure, of the presence of catalysts. Also nuclear reactions of given radioisotope cannot be slowed down, speeded up, or stopped.

7 0

3 years ago