Write the symbol for the element which has the smallest atomic radius.c, al, li, ne.

2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.

azamat

Answer:

$K=1.12x10^9$

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

$K=\frac{[NO_2]^2}{[NO]^2[O_2]}$

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

$K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9$

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0

3 years ago

With a well labeled diagram explain the stages of meiosis and mitosis

NNADVOKAT [17]

Answer:

hope that helps

7 0

3 years ago

When two atoms of element Y combine, Y2 molecule is formed.

Liula [17]

Answer:

chemical bond

Explanation:

because this is the only bond used in chemical reactions just like molecules of atoms are combined by this bond

5 0

3 years ago

Which solution has a higher percent ionization of the acid, a 0.10M solution of HC2H3O2(aq) or a 0.010M solution of HC2H3O2(aq)

Svetach [21]

Answer:

The solution 0.010 M has a higher percent ionization of the acid.

Explanation:

The percent ionization can be found using the following equation:

$\% I = \frac{[H_{3}O^{+}]}{[CH_{3}COOH]} \times 100$

Since we know the acid concentration in the two cases, we need to find [H₃O⁺].

By using the dissociation of acetic acid in the water we can calculate the concentration of H₃O⁺ in the two cases:

1. Case 1 (0.1 M):

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺(aq) (1)

0.1 - x x x

$Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}$ (2)

Where:

Ka: is the dissociation constant of acetic acid = 1.7x10⁻⁵.

$1.7 \cdot 10^{-5} = \frac{x^{2}}{0.1 - x}$

$1.7 \cdot 10^{-5}*(0.1 - x) - x^{2} = 0$

By solving the above equation for x we have:

x = 1.29x10⁻³ M = [CH₃COO⁻] = [H₃O⁺]

Hence, the percent ionization is:

$\% I = \frac{1.29 \cdot 10^{-3} M}{0.1 M} \times 100 = 1.29 \%$

2. Case 2 (0.01 M):

The dissociation constant from reaction (1) is:

$Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}$

With [CH₃COOH] = 0.01 M

$1.7 \cdot 10^{-5} = \frac{x^{2}}{0.01 - x}$

$1.7 \cdot 10^{-5}*(0.01 - x) - x^{2} = 0$

By solving the above equation for x:

x = 4.04x10⁻⁴ M = [CH₃COO⁻] = [H₃O⁺]

Then, the percent ionization for this case is:

$\% I = \frac{4.04 \cdot 10^{-4} M}{0.01 M} \times 100 = 4.04 \%$

As we can see, the solution 0.010 M has a higher percent ionization of the acetic acid.

Therefore, the solution 0.010 M has a higher percent ionization of the acid.

I hope it helps you!

4 0

3 years ago

The element with the smallest iconic radius in the alkaline earth metals is

olya-2409 [2.1K]

Answer:

Metal.

Explanation:

Metal has the smallest iconic radius in the alkaline earth metal.

8 0

3 years ago