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3 years ago

8

When calcium forms an ion that has the same number of outer electrons as which noble gas?

1 answer:

Talja [164]3 years ago

6 0

The answer is 4 krypton

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Radioactive isotopes have been used commercially in:

Law Incorporation [45]

<span>Radioactive isotopes have been used commercially in all these applications.
The last option (D) is your answer

</span>

3 0

3 years ago

Water contains 7.0 mg/L of soluble ion (Fe2 ) that is to be oxidized by aeration to a concentration of 0.25 mg/L. The pH of the

SpyIntel [72]

Answer:

For CMFR:

V = 5286 m³

t = 2.6 hr

For PFR:

V = 529 m³

t = 19 minutes

Explanation:

Find attached for the calculation.

4 0

4 years ago

A. O CN<br> B. O NH4+<br> C. O s2-<br> D. O Cro 2-<br> Which of the following is an oxyanion?

ANEK [815]

C i think i got this off of usa test prep in class

5 0

4 years ago

1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?

Ratling [72]

Hello!

1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

<u><em>We have the following data:</em></u>

Vo (initial volume) = 1.00 L

V (final volume) = 473 mL → 0.473 L

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)

P (final pressure) = ? (in atm)

<u><em>We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:</em></u>

$P_0*V_0 = P*V$

$1*1 = P*0.473$

$1 = 0.473\:P$

$0.473\:P = 1$

$P = \dfrac{1}{0.473}$

$\boxed{\boxed{P \approx 2.11\:atm}}\:\:\:\:\:\:\bf\green{\checkmark}$

<u><em>Answer: </em></u>

<u><em>The new pressure of the gas is 2.11 atm </em></u>

___________________________________

$\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

3 0

3 years ago

The decomposition of nitrogen dioxide to nitrogen monoxide and oxygen gas is a second order process as suspected in the previous

lesya [120]

Explanation:

For the given reaction $2NO_{2} \rightarrow 2NO + O_{2}$

Now, expression for half-life of a second order reaction is as follows.

$t_{1/2} = \frac{1}{[A_{0}]k}$ ....... (1)

Second half life of this reaction will be $t_{1/4}$ . So, expression for this will be as follows.

$t_{1/4}$ = $\frac{1}{k} [\frac{1}{[A]_{f}} - \frac{1}{[A_{0}]}]$ ...(2)

where $[A]_{f}$ is the final concentration that is, $\frac{[A]_{0}}{4}$ here and $[A]_{i}$ is the initial concentration.

Hence, putting these values into equation (2) formula as follows.

$t_{1/4}$ = $\frac{1}{k} [\frac{4}{[A]_{0}} - \frac{1}{[A_{0}]}]$

= $\frac{3}{[A_{0}]k}$ ...... (3)

Now, dividing equation (3) by equation (1) as follows.

$\frac{t_{1/4}}{t_{1/2}}$ = $\frac{3}{[A_{0}]k} \times [A_{0}]k$

= 3

or, $t_{1/4}$ = 3 $t_{1/2}$

Thus, we can conclude that one would expect the second half-life of this reaction to be three times the first half-life of this reaction.

6 0

3 years ago