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3 years ago

The air pressure inside a submarine is 0.832atm. What would the pressure be in mnhg

Chemistry

1 answer:

Gwar [14]3 years ago

5 0

Answer:

632.32 mmHg

Explanation:

Millimetre mercury:

It is the monometric unit of pressure. It is define as "The pressure exerted by the column pf mercury at the height of 1 millimetre.

It is represented as mmHg.

It can also be written as mm Hg.

Atmosphere (atm):

It is barometric pressure, define as "The force exerted by atmospheric column on per unit area".

It is written as "atm".

Conversion of atm to mmHg:

0.832 atm × 760 mmHg / 1 atm

632.32 mmHg

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How many milliliters of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride (FM 121.135) to give a pH of 7.60 in a fina

liubo4ka [24]

Answer:

41.64mL of NaOH 0.500M must be added to obtain the desire pH

Explanation:

It is possible to find pH of a buffer by using H-H equation, thus:

pH = pka + log [A⁻] / [HA]

Where [HA] is concentration of the weak acid TRIS-HCl and [A⁻] is concentration of its conjugate acid.

Replacing in H-H equation:

7.60 = 8.072 + log [A⁻] / [HA]

0.3373 = [A⁻] / [HA] (1)

10.0g of TRIS-HCl (Molar mass: 121.135g/mol) are:

10.0g ₓ (1mol / 121.135g) = 0.08255 moles of acid. That means moles of both the acid and conjugate base are:

[A⁻] + [HA] = 0.08255 (2)

Replacing (1) in (2):

0.3373 = 0.08255 - [HA] / [HA]

0.3373[HA] = 0.08255 - [HA]

1.3373[HA] = 0.08255

[HA] = 0.06173 moles

Thus:

[A⁻] = 0.08255 - 0.06173 = 0.02082 moles [A⁻]

The moles of A⁻ comes from the reaction of the weak acid with NaOH, that is:

HA + NaOH → A⁻ + H₂O + K⁺

Thus, you need to add 0.02082 moles of NaOH to produce 0.02082 moles of A⁻. As NaOH solution is 0.500M:

0.02082 moles NaOH ₓ (1L / 0.500mol) = 0.04164L of NaOH 0.500M =

<h3>41.64mL of NaOH 0.500M must be added to obtain the desire pH</h3>

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3 years ago

What is the empirical formula of a compound that is 24.42 % calcium, 17.07 % nitrogen, and 58.5% oxygen?

REY [17]

Answer:

$CaN_{2} O_{6}$

Explanation:

When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.

So:

24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O

The next step is to divide each mass by their molar mass to convert your grams to moles.

24.42/40.08 = 0.6092 mol

17.07/14.01 = 1.218 mol

58.85/15.99 = 3.680 mol

Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.

0.6092 mol/0.6092 mol = 1

1.218 mol/0.6092 mol = 2

3.680 mol/0.6092 mol = 6

So the empirical formula is $CaN_{2} O_{6}$

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