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3 years ago

What is the oxidation state of nitrogen in nano2

Chemistry

1 answer:

TiliK225 [7]3 years ago

3 0

Answer:

Therefore, the oxidation state of N in NaNO₂ is +3

Explanation:

Problem: calculating the oxidation state of nitrogen N in NaNO₂

let us denote the oxidation number of Nitrogen as N:

We know from the periodic table that Na has an oxidation number of +1 i.e it will readily want to lose 1 electron so as to complete its octet.

Oxygen is known to have an oxidation number of -2

Summation of the oxidation number of each atoms is 0 for neutral compound.

Therefore to calculate the oxidation state of Nitrogen in NaNO₂, we express as:

+1 + N + (-2 x 2) = 0

1 + N = 4

N = 4-1 = +3

Therefore, the oxidation state of N in NaNO₂ is +3

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Which of the molecules represented below contains carbon with sp2 hybridization?

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Answer:

Explanation:

sp² hybridization is found in those compounds having double bond .

Out of the given compounds only C₂H₂Cl₂ has double bond so this compound contains carbon with sp² hybridization .

Rest have sp³ hybridization because they are saturated compounds .

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3 years ago

12. Describe the results of a chemical change. List four<br> indicators of chemical change.

marshall27 [118]

Answer:

1. Color Change

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3 years ago

Be sure to answer all parts. Consider the reaction A + B → Products From the following data obtained at a certain temperature, d

worty [1.4K]

Answer : The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B\rightarrow Products$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first observation:

$3.20\times 10^{-1}=k(1.50)^a(1.50)^b$ ....(1)

Expression for rate law for second observation:

$3.20\times 10^{-1}=k(1.50)^a(2.50)^b$ ....(2)

Expression for rate law for third observation:

$6.40\times 10^{-1}=k(3.00)^a(1.50)^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(1.50)^a(2.50)^b}{k(1.50)^a(1.50)^b}\\\\1=1.66^b\\b=0$

Dividing 1 from 3, we get:

$\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(3.00)^a(1.50)^b}{k(1.50)^a(1.50)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[A]^1[B]^0$

$\text{Rate}=k[A]$

Thus,

The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

7 0

3 years ago

You need to prepare at 2.0 mL sample of a diluted drug for injection. The total amount of the drug to be injected in this 2.0 mL

pav-90 [236]

Answer:

a) The concentration of drug in the bottle is 9.8 mg/ml

b) 0.15 ml drug solution + 1.85 ml saline.

c) 4.9 × 10⁻⁵ mol/l

Explanation:

Hi there!

a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml

b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:

0.0210 mg/kg-body mass * 70 kg = 1.47 mg drug.

The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)

If 9.8 mg of the drug is contained in 1 ml of solution, then 1.47 mg drug will be present in (1.47 mg * 1 ml/ 9.8 mg) 0.15 ml.

To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline

c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.

Let´s convert it to molarity:

0.735 mg/ml * 1000 ml/l * 0.001 g/mg* 1 mol/ 15000 g = 4.9 × 10⁻⁵ mol/l

3 0

3 years ago

A student was assigned the task of determining the identity of an unknown liquid. The student weighed a clean, dry 250-mL Erlenm

LiRa [457]

Answer:

248.4 mL

Explanation:

Erlenmeyer = 78.649 g

Erlenmeyer + Water = 327.039 g

Water = (Erlenmeyer + Water) - Erlenmeyer

Water = 327.039 - 78.649

Water = 248.4 g

if the density of water is 1 g/mL, we can say that each mL of water weigh 1 g, so we have 248.4 mL of water in the Erlenmeyer Flask.

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3 years ago