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IRISSAK [1]
3 years ago
10

If a gas has a volume of 1000 ML at a temperature of 23°C and a pressure of 100 mmhg, what is it’s volume under standard conditi

ons of temperature and pressure?
Chemistry
1 answer:
Colt1911 [192]3 years ago
8 0

Answer:

119.7 mL.

Explanation:

  • From the general law of ideal gases:

<em>PV = nRT.</em>

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas (K).

  • For the same no. of moles of the gas at two different (P, V, and T):

<em>P₁V₁/T₁ = P₂V₂/T₂.</em>

  • P₁ = 100.0 mmHg, V₁ = 1000.0 mL, T₁ = 23°C + 273 = 296 K.
  • P₂ = 1.0 atm = 760.0 mmHg (standard P), V₂ = ??? mL, T₂ = 0.0°C + 273 = 273.0 K (standard T).

<em>∴ V₂ = (P₁V₁T₂)/(T₁P₂) </em>= (100.0 mmHg)(1000.0 mL)(273.0 K)/(296 K)(760.0 mmHg) =  121.4 <em>mL.</em>

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1.5 moles of Al is how many<br> atoms?
aev [14]

Answer:

<h2>9.03 × 10²³ atoms </h2>

Explanation:

The number of atoms of Al can be found by using the formula

<h3>N = n × L</h3>

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

N = 1.5 × 6.02 × 10²³

We have the final answer as

<h3>9.03 × 10²³ atoms</h3>

Hope this helps you

4 0
2 years ago
What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as one made from 475 mL of 0.200 M
Fed [463]

<u>36 ml of NaOh and</u><u> 464 ml</u><u> of </u><u>HCOOH</u><u> would be enough to form 500 ml of a buffer with the same pH as the buffer made with </u><u>benzoic acid </u><u>and NaOH.</u>

What is benzoic acid found in?

  • Some natural sources of benzoic acid include: ​Fruits:​ Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
  • ​Spices:​ Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.

Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.

Amount of moles of NaOH -2 × 0.025 =  0.05 mol

Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol

In this case, we can calculate the pH produced by the buffer of these two reagents, as follows

pH = pKa + log\frac{base}{acid}

4.2 + log\frac{0.05}{0.045} = 4.245

We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows

pH = pKa + log\frac{base}{acid}

4.245 = 3.75 + log\frac{base}{acid}

log\frac{base}{acid} = 0.5

\frac{base}{acid} = 3.162

Now we must solve the equation above. This will be done using the following values

\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 0.464 L

With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.

NaOH volume

( 0.5 - 0.464)L

0.036L .................... 36ml

HCOOH volume

500 - 36 = 464mL

Learn more about benzoic acid

brainly.com/question/24052816

#SPJ4

3 0
1 year ago
Difference between sodium ion and sodium atom.<br>(give in points not in paragraph)​
Natali5045456 [20]

Answer:

Explanation:

SODIUM ATOM;

SODIUM ATOM IS NEUTRAL

SODIUM ION;

IT IS A CHARGED SPECIE WITH A CHARGE OF +1

SODIUM ATOM:

THE NUMBER OF PROTONS AND ELECTRONS ARE SAME ie:11

SODIUM ION:

NUMBER OF PROTONS AND ELECTRONS ARE NOT SAME ie. ELETRON: 10, PROTONS:11

HOPE IT WILL HELP:)

3 0
3 years ago
What is the new volume of a balloon originally at 755 torr and 5.00 l is placed in a container in which the pressure is increase
Luba_88 [7]

Answer is: volume will be 3.97 liters.

Boyle's Law: the pressure volume law -  volume of a given amount of gas held  varies inversely with the applied pressure when the temperature and mass are constant.

p₁V₁ = p₂V₂.

p₁ = 755 torr.

V₁ = 5.00 l.

p₂ = 1.25 atm · 760 torr/atm.

p₂ = 950 torr.

755 torr · 5 l = 950 torr · V₂.

V₂ = 755 torr · 5 l / 950 torr.

V₂ =  3.97 l.

When pressure goes up, volume goes down.

When volume goes up, pressure goes down.

3 0
3 years ago
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If 0.4506 g of KHP standard were titrated with NaOH and the endpoint was reached at 27.96 mL, what is the concentration (molarit
Lemur [1.5K]
Answer is in attachment.

Enjoy :)

6 0
3 years ago
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