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Anuta_ua [19.1K]
3 years ago
8

2. At STP, a sample of gas occupies 24.5 mL. Calculate the volume of this gas at a

Chemistry
1 answer:
Law Incorporation [45]3 years ago
6 0

Answer:

11.7 ml = V

Explanation: Here

You might be interested in
Aqueous concentrated nitric acid is 69% hno3 by weight and has a density of 1.42 g/ml.
OleMash [197]

Answer: -

15.55 M

35.325 molal

Explanation: -

Let the volume of the solution be 1000 mL.

Density of nitric acid = 1.42 g/ mL

Total Mass of nitric acid Solution = Volume of nitric acid x Density of nitric acid

= 1000 mL x 1.42 g/ mL

= 1420 g.

Percentage of HNO₃ = 69%

Amount of HNO₃ = \frac{69} {100} x 1420 g

= 979.8 g

Molar mass of HNO₃ = 1 x 1 + 14 x 1 + 16 x 3 = 63 g /mol

Number of moles of HNO₃ = \frac{979.8 g}{63 g/ mol}

= 15.55 mol

Molarity is defined as number of moles per 1000 mL

We had taken 1000 mL as volume and found it to contain 15.55 moles.

Molarity of HNO₃ = 15.55 M

Mass of water = Total mass of nitric acid solution - mass of nitric acid

= 1420 - 979.8

= 440.2 g

So we see that 440.2 g of water contains 15.55 moles of HNO₃

Molality is defined as number of moles of HNO₃ present per 1000 g of water.

Molality of HNO₃ = \frac{15.55 x 1000}{440.2}

= 35.325 molal

3 0
3 years ago
Let this be an easy 5 points for your brainly account all is ask for in return is a "thank you"
Svet_ta [14]

Answer:

The number of neutrons present in one atom of isotope of Silicon of mass 28 amu is<u> 14 neutrons</u>

Explanation:

Symbol of Si isotope

_{14}^{28}\textrm{Si}

<u>Number of Neutron = Mass number - Atomic Number</u>

Mass number = Total number of protons and neutrons present in the nucleus of the atom.For Si = 28 amu

Atomic Number = Total number of Protons present in the nucleus.

Si = 14

Number of neutron = 24 - 14

                                = 14

4 0
3 years ago
Four cars travel at the same speed down a road. Their mass and kinetic energy are shown in the graph,
Alex777 [14]

Answer:B

Explanation:

3 0
2 years ago
Which one is the catalyst?
yan [13]

Answer:

the catalyst is the two gray dots

5 0
3 years ago
Dilution question In many of the experiments, you will be asked to prepare a standard solution by diluting a stock solution. You
riadik2000 [5.3K]

Answer:

We could do two 1:50 dilutions and one 1:4 dilutions.

Explanation:

Hi there!

A solution that is 1000 ug/ ml  (or 1000 mg / l) is 1000 ppm.

Knowing that 1 ppm = 1000 ppb, 100 ppb is 0.1 ppm.

Then, we have to dilute the stock solution (1000 ppm / 0.1 ppm) 10000 times.

We could do two 1:50 dilutions and one 1:4 dilutions (50 · 50 · 4 = 10000). Since the first dilution is 1:50, you will use the smallest quantity of the stock solution (if we use the 10.00 ml flask):

First step (1:50 dilution):

Take 0.2 ml of the stock solution using the third dispenser (20 - 200 ul), and pour it in the 10.00 ml flask. Fill with water to the mark (concentration : 1000 ppm / 50 = 20 ppm).

Step 2 (1:50 dilution):

Take 0.2 ml of the solution made in step 1 and pour it in another 10.00 ml flask. Fill with water to the mark. Concentration 20 ppm/ 50 = 0.4 ppm)

Step 3 (1:4 dilution):

Take 2.5 ml of the solution made in step 3 (using the first dispenser 1 - 5 ml) and pour it in a 10.00 ml flask. Fill with water to the mark. Concentration 0.4 ppm / 4 = 0.1 ppm = 100 ppb.

6 0
3 years ago
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