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3 years ago

The melting point of h2o is 0 degrees celsius. this is the same as its:

Chemistry

1 answer:

professor190 [17]3 years ago

6 0

This is the same as its freezing point

hope this helps

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How many liters of volume is one mole of gas at standard temperature and<br> pressure?

horsena [70]

22.7 liters

The molar volume of an ideal gas depends on the temperature and pressure. One mole of any ideal gas occupies 22.7 liters at 0 0C and 1 bar (STP).

Hope this helped

3 0

2 years ago

The monovalent salt concentration (the predominant solute in the blood cell) for a sample of red blood cells is 0.13 moles/liter

11111nata11111 [884]

Answer:

The osmotic pressure of cell is $648.3$ KPa

Explanation:

As we know the osmotic pressure is equal to

$\pi = icRT$

Where

i is the Van Hoff factor

c is the concentration of solution

R is the ideal gas constant

and T is the temperature.

Substituting the given values, we get -

$\pi = 2 * 0.13 * 0.08206 * 300\\$

$\pi = 648.3$ KPa

4 0

2 years ago

Magnesium reacts steadily with hydrochloric acid

balu736 [363]

Magnesium + Hydrocloric acid -> Magnesium chloride + hydrogen

You can observe a single displacement reaction

"Describe to show that the has formed is hydrogen"

I don't know what you mean. I can show the chemical equation though.

Mg(s) + 2 HCl(aq) --> MgCl 2(aq) + H 2(g)

3 0

3 years ago

Based on the concept of the global conveyor belt, what happens to ocean water as it moves from Antarctica to the equator?

Korolek [52]

Answer:

I think the answer is C

Explanation:

5 0

2 years ago

Read 2 more answers

A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th

Mashcka [7]

Explanation:

It is known that $K_{a}$ of $HNO_{2}$ = $4.5 \times 10^{-4}$ .

(a) Relation between $K_{a}$ and $pK_{a}$ is as follows.

$pK_{a} = -log (K_{a})$

Putting the values into the above formula as follows.

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Also, relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log\frac{[conjugate base]}{[acid]}$

= $3.347+ log \frac{0.15}{0.12}$

= 3.44

Therefore, pH of the buffer is 3.44.

(b) No. of moles of HCl added = $Molarity \times volume$

= $11.6 M \times 0.001 L$

= 0.0116 mol

In the given reaction, $NO^{-}_{2}$ will react with $H^{+}$ to form $HNO_{2}$

Hence, before the reaction:

No. of moles of $NO^{-}_{2}$ = $0.15 M \times 1.0 L$

= 0.15 mol

And, no. of moles of $HNO_{2}$ = $0.12 M \times 1.0 L$

= 0.12 mol

On the other hand, after the reaction :

No. of moles of $NO^{-}_{2}$ = moles present initially - moles added

= (0.15 - 0.0116) mol

= 0.1384 mol

Moles of $HNO_{2}$ = moles present initially + moles added

= (0.12 + 0.0116) mol

= 0.1316 mol

As, $K_{a}$ = $4.5 \times 10^{-4}$

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = $pK_{a} + log \frac{[conjugate base]}{[acid]}$

= 3.347+ log {0.1384/0.1316}

= 3.369

= 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0

3 years ago