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3 years ago

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Chemistry

1 answer:

notsponge [240]3 years ago

6 0

Answer:

gime fre points

Explanation:

fre points for me!

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According to rounding rules for addition the sum of 27.1, 34.538, and 37.68 is

Katena32 [7]

The answer should be...99.318!

5 0

2 years ago

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:

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

5 0

3 years ago

Polonium is a rare element with 33 radioisotopes. The most common one, 210Po, has 84 protons and 126 neutrons. When 210Po decays

Fofino [41]

Answer:

Lead Pb

Explanation:

Firstly, we need to know what occurs when a radioisotope emits an alpha particle. An alpha particle is an helium atom. When an isotope emits an alpha particle, it loses an helium atom corresponding to subtracting 4 from its mass number and 2 from its atomic number. This of course coupled with the release of radiation.

Now, we polonium has a proton number of 84 and a mass number of 210. Subtracting 2 and 4 respectively from its proton and mass numbers will yield 82 and 206 proton and mass numbers respectively.

Hence, the decomposition of the Po-210 isotope will yield an element with 82 proton number and 206 mass number. This corresponds to the element Lead.

210Po ——> 206Pb + alpha particle + radiation

6 0

2 years ago

Part of Dr. Ritchey’s work in her previous job involved boiling and condensing hydrocarbon mixtures. In one of her experiments,

Hitman42 [59]

Answer:

what is the inlet concentration of n-pentane in % by volume? = 34.4%

Explanation:

check below for explanation in the attachment