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4 years ago

Consider the total ionic equation below.

Chemistry

1 answer:

Rzqust [24]4 years ago

3 0

Answer:

The correct answer is $CrO^{2-} _{4}$ and $Ba^{2+}$

Explanation:

A "spectator ion" is an ion that exists as a reagent and as a product in a chemical reaction. This ion has the particularity of not affecting the equilibrium.

In this case, the two components that do not see their charges modified on both sides of the reaction are $CrO^{2-} _{4}$ and $Ba^{2+}$

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What is the energy of a photon that emits a light of frequency 6.42 x 1014 Hz?<br><br>

svetlana [45]

Answer:

Option B. 4.25×10¯¹⁹ J

Explanation:

From the question given above, the following data were obtained:

Frequency (f) = 6.42×10¹⁴ Hz

Energy (E) =?

Energy and frequency are related by the following equation:

Energy (E) = Planck's constant (h) × frequency (f)

E = hf

With the above formula, we can obtain the energy of the photon as follow:

Frequency (f) = 6.42×10¹⁴ Hz

Planck's constant (h) = 6.63×10¯³⁴ Js

Energy (E) =?

E = hf

E = 6.63×10¯³⁴ × 6.42×10¹⁴

E = 4.25×10¯¹⁹ J

Thus, the energy of the photon is 4.25×10¯¹⁹ J

3 0

3 years ago

Read 2 more answers

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

Initial concentration of $\rm HA$ without any dissociation:

$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .