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2 years ago

Calculate the pH of a (0.000765) M solution of KOH

Chemistry

1 answer:

likoan [24]2 years ago

6 0

Answer:

pH = - log(0.000765)

= -(-3.11)

= 3.1

so the solution is basic

rest you can check values using calculator

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Which of these electron transitions correspond to absorption of energy and which to emission?

Keith_Richards [23]

The electron transitions representing absorption of energy are:

(a) n = 2 to n = 4

(d) n = 3 to n = 4

and those which represents emissions of energy are:

(b) n = 3 to n = 1

An electron emits energy when it jumps from a higher energy level to a lower energy level transitions.

An electron absorbs energy to jump from a lower energy level to a higher energy level transitions.

(a) n = 2 to n = 4

Here, the initial level (2) is lower than the final level (4).

Hence energy is absorbed.

(b) n = 3 to n = 1

Here, the initial level (3) is higher than the final level (1).

Hence energy is released.

Here, the initial level (5) is higher than the final level (2).

Hence energy is released.

(d) n = 3 to n = 4

Here, the initial level (3) is lower than the final level (4).

Hence energy is absorbed.

The question is incomplete. Find the complete question here:

Which of these electron transitions correspond to absorption of energy and which to emission?

(a) n = 2 to n = 4

(b) n = 3 to n = 1

(d) n = 3 to n = 4

Learn more about electron transitions at brainly.com/question/2079874

#SPJ4

4 0

1 year ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

2 years ago

An example of a colloid which is composed of a liquid dispersed in a gas would be

Aleksandr-060686 [28]

Any colloid consisting of a solid dispersed in a gas is called a smoke. A liquid dispersed in a gas is referred to as a fog. So the answer would be smoke

6 0

2 years ago

In the name carbon dioxide what does the prefix of the second word indicate about a molecule of a dioxide A. It has two carbon a

brilliants [131]

Answer:

The answer is option B, that is, it has two oxygen atoms.

5 0

2 years ago

Baking soda (nahco3) is an active ingredient in some antacids. determine the percent composition of oxygen in baking soda.

telo118 [61]

Assuming the question is asking for percent composition by mass,

The molar mass of sodium bicarbonate is:

22.990 + 1.008 + 12.011 + 3*15.999 = 84.006 g/mol

The molar mass of 3 oxygen atoms is:

3*15.999 = 47.997 g/mol

The percentage composition by mass is:

47.997/84.006 * 100% = 57.1% (3 s.f.)

8 0

2 years ago