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4 years ago

Is the following chemical equation balanced or unbalanced? Balanced or Unbalanced

Chemistry

2 answers:

ale4655 [162]4 years ago

4 0

Answer:

This equation is balanced.

Explanation:

To determine whether an equation is balanced or not, we must make sure that there are the same number of each type of atoms in both the reactants (on the left side of the equation) and the products (on the right side of the equation).

We can see that there are two atoms of nitrogen on the left side of the equation. When we look at the right side of the equation, we see that there are two molecules of NH6, with one atom of nitrogen each. Since 2 * 1 = 2, there are two atoms of nitrogen in the products as well. This means that the nitrogen atoms are balanced.

Next, we must look at the hydrogen atoms. There are 6 molecules of H2, which contains 2 atoms of hydrogen each. Since 6 * 2 = 12, there are 12 atoms of hydrogen in the reactants. On the right side of the equation, there are 2 molecules of NH6, which contains 6 atoms of hydrogen each. Since 2 * 6 = 12, there are 12 atoms of hydrogen in the products as well. Since there are 12 atoms on both sides of the equation, the hydrogen atoms are balanced.

Since both types of atoms that are presented in this equation are balanced, we can conclude that the overall equation is balanced.

Hope this helps!

xenn [34]4 years ago

3 0

Answer:

Unbalanced

Explanation:

A balanced chemical equation must satisfy three conditions. It must:

Give the correct formulas for all reactants and products.
Have the same number of atoms of each element on both sides of the reaction arrow (balance mass).
Have the same net charge on each side of the reaction arrow (balance charge).

Your equation is unbalanced because it fails the first condition.

It satisfies the second condition, because there are 2 N atoms and 12 H atoms on each side of the arrow.

It also satisfies the third condition, because there are no charges on either side of the arrow.

However, a correctly balanced chemical equation must satisfy all three conditions.

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Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:

posledela

Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Explanation:

Equilibrium concentration of $HNO_3$ = $\frac{15.5g}{63g/mol\times 9.5L}=0.026M$

Equilibrium concentration of $NO$ = $\frac{16.6g}{30g/mol\times 9.5L}=0.058M$

Equilibrium concentration of $NO_2$ = $\frac{22.5g}{46g/mol\times 9.5L}=0.051M$

Equilibrium concentration of $H_2O$ = $\frac{189.0g}{18g/mol\times 9.5L}=1.10M$

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

For the given chemical reaction:

$2HNO_3(aq)+NO(g)\rightarrow 3NO_2(g)+H_2O(l)$

The expression for $K_c$ is written as:

$K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}$

$K_c=\frac{(0.051)^3\times (1.10)^1}{(0.026)^2\times (0.058)^1}$

$K_c=3.72$

Thus the value of the equilibrium constant Kc for this reaction is 3.72

5 0

3 years ago

A city continuously disposes of effluent from a wastewater treatment plant into a river. The minimum flow in the river is 130 m3

Vera_Pavlovna [14]

Answer:

$2.54\ \text{mg/L}$

Explanation:

C = Allowable concentration = 1.1 mg/L

$Q_1$ = Flow rate of river = $130\ \text{m}^/\text{s}$

$Q_2$ = Discharge from plant = $37\ \text{m}^3/\text{s}$

$C_1$ = Background concentration = 0.69 mg/L

$C_2$ = Maximum concentration that of the pollutant

The concentration of the mixture will be

$C=\dfrac{Q_1C_1+Q_2C_2}{Q_1+Q_2}\\\Rightarrow C_2=\dfrac{C(Q_1+Q_2)-Q_1C_1}{Q_2}\\\Rightarrow C_2=\dfrac{1.1(130+37)-130\times 0.69}{37}\\\Rightarrow C_2=2.54\ \text{mg/L}$

The maximum concentration that of the pollutant (in mg/L) that can be safely discharged from the wastewater treatment plant is $2.54\ \text{mg/L}$ .

6 0

3 years ago

12 Significant Figures in Addition and

Pavlova-9 [17]

Answer: 85.6

Explanation:

8 0

3 years ago

Which choice gives the correct oxidation numbers for all three elements in rb2so3 in the order that the elements are shown in th

vlabodo [156]

<h3>Answer:</h3>

Rb = + 1

S = + 4

O = - 2

<h3>Explanation:</h3>

Oxidation states of the elements were calculated keeping in mind the basic rules of assigning oxidation states which included assignment of +1 charge to first group elements i.e. Rubidium (Rb) and assignment of -2 charge to Oxygen atom. Then the oxidation state of Sulfur was calculated as follow,

Rb₂ + S + O₃ = 0

Above zero (0) means that the overall molecule is neutral.

Putting values of Rb and O,

(+1)₂ + S + (-2)₃ = 0

(+2) + S + (-6) = 0

+2 + S - 6 = 0

S - 6 = -2

S = -2 + 6

S = + 4

7 0

3 years ago

The K w for water at 0 ∘ C is 0.12 × 10 − 14 . Calculate the pH of a neutral aqueous solution at 0 ∘ C.

dusya [7]

Answer:

pH = 7.46

Explanation:

2H₂O ⇄ H₃O⁺ . OH⁻ Kw = [H₃O⁺] . [OH⁻]

[H₃O⁺] = [OH⁻]

√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M

- log [H₃O⁺] = pH

- log 3.46×10⁻⁸ = pH → 7.46

6 0

4 years ago