All categories

3 years ago

in the reaction mgcl2 + 2koh mg(oh)2 + 2kcl, if 3 moles mgcl2 are added to 4 moles koh what determines how much mg(oh)2 is made

2 answers:

6 0

Every mole of MgCl2 reacts with 2 moles of KOH, therefore the 4 moles of KOH will only react with 2 moles of MgCl2, making it the limiting reagent and therefore KOH determines how much Mg(OH)2 is produced.

yanalaym [24]3 years ago

3 0

The amount of KOH --apex

You might be interested in

In a voltaic cell, the ____

alekssr [168]

In a voltaic cell , the anode electrons and is oxidized , while the cathode electrons and is reduced.

5 0

3 years ago

Read 2 more answers

The overall electrical charge on a stable element is_____

Elodia [21]

Answer: the answer is 0

Explanation:A stable atom has a net charge of 0. In other words, it has an equal number of protons and electrons. The positive protons cancel out the negative electrons.

7 0

3 years ago

A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to

Snezhnost [94]

Explanation:

It is known that $pK_{a}$ of propionic acid = 4.87

And, initial concentration of propionic acid = $\frac{0.19}{1.20}$

= 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

= 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence, pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.216}{0.158}$

= 4.87 + log (1.36)

= 5.00

Therefore, when 0.02 mol NaOH is added then,

Moles of propionic acid = 0.19 - 0.02

= 0.17 mol

Hence, concentration of propionic acid = $\frac{0.17}{1.20 L}$

= 0.14 M

and, moles of sodium propionic acid = (0.26 + 0.02) mol

= 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

$\frac{0.28 mol}{1.20 L}$

= 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.23}{0.14}$

= 4.87 + log (1.64)

= 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

Therefore, when 0.02 mol HI is added then,

Moles of propionic acid = 0.19 + 0.02

= 0.21 mol

Hence, concentration of propionic acid = $\frac{0.21}{1.20 L}$

= 0.175 M

and, moles of sodium propionic acid = (0.26 - 0.02) mol

= 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

$\frac{0.24 mol}{1.20 L}$

= 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.2}{0.175}$

= 4.87 + log (0.114)

= 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0

3 years ago

Which is a Characteristic of a non elecrolyte

myrzilka [38]

Answer:

A nonelectrolyte is a substance that does not exist in an ionic form in aqueous solution. Nonelectrolytes tend to be poor electrical conductors and don't readily dissociate into ions when melted or dissolved. Solutions of nonelectrolytes do not conduct electricity.

8 0

3 years ago

It is thought that Vitamin C might be an effective preventative

Anon25 [30]

Answer:

IF the fruit juice contains a high level of vitamin C, THEN the preventative effectiveness against common cold increases.

Explanation:

The hypothesis is a testable explanation of a scientific investigation. It aims at predicting the outcome of the experiment. One feature of the hypothesis is that it must be testable. The hypothesis is usually written in an "IF, THEN" format.

This question is regarding an experiment to test the amount of vitamin C in fruit juice. The vitamin C is thought to be an effective preventative against common cold. Hence, the hypothesis connects the effect on common cold (dependent variable) with the amount of vitamin C (independent variable). The hypothesis can be written as:

IF the fruit juice contains a high level of vitamin C, THEN the preventative effectiveness against common cold increases.

7 0

3 years ago