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VLD [36.1K]
3 years ago
15

Help me pls!! List all three sets of opposing forces.

Chemistry
1 answer:
jek_recluse [69]3 years ago
7 0

Answer:

System-wide consistency versus institutional autonomy, efficient versus effective assessment, and promotion of student progression versus enforcement of academic standards.

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Commercial hydrochloric acid (HCl) is typically labeled as being 38.0 % (weight %). The density of HCl is 1.19 g/mL. a) What is
Strike441 [17]

<u>Answer:</u>

<u>For a:</u> The molarity of commercial HCl solution is 12.39 M.

<u>For b:</u> The molality of commercial HCl solution is 16.79 m.

<u>For c:</u> The volume of commercial HCl solution needed is 2.42 L.

<u>Explanation:</u>

We are given:

Mass % of commercial HCl solution = 38 %

This means that 38 grams of HCl is present in 100 grams of solution.

To calculate the volume of solution, we use the equation:

Density=\frac{Mass}{Volume}

Density of HCl solution = 1.19 g/mL

Mass of solution = 100 g

Putting values in above equation:

1.19g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=84.034mL

  • <u>For a:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = ?

Molar mass of HCl = 36.5 g/mol

Volume of solution = 84.034 mL

Mass of HCl = 38 g

Putting values in above equation, we get:

\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 84.034}\\\\\text{Molality of commercial HCl solution}=12.39M

Hence, the molarity of commercial HCl solution is 12.39 M.

  • <u>For b:</u>

To calculate the molality of solution, we use the equation:

Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m_{solute} = Given mass of solute (HCl) = 38 g

M_{solute} = Molar mass of solute (HCl) = 36.5 g/mol

W_{solvent} = Mass of solvent = 100 - 38 = 62 g

Putting values in above equation, we get:

\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 62}\\\\\text{Molality of commercial HCl solution}=16.79m

Hence, the molality of commercial HCl solution is 16.79 m.

  • <u>For c:</u>

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated solution

M_2\text{ and }V_2 are the molarity and volume of diluted solution

We are given:

M_1=6M\\V_1=5.00L\\M_2=12.39M\\V_2=?L

Putting values in above equation, we get:

6\times 5=12.39\times V_2\\\\V_2=2.42L

Hence, the volume of commercial HCl solution needed is 2.42 L.

5 0
3 years ago
Which of the following sets represents a pair of isotopes?
Maslowich
The set that represent a pair of isotope is 9B AND 11B.
Isotopes are chemical element which have two or more forms, these forms have the same number of electrons but different number of neutrons.
In an atom of an element, the nucleus contains the neutron and the proton. For a neutral atom, the number of proton and neutron are always the same, while the number of neutron differ. Thus, for the correct option, B stands for the element Boron, which has atomic number of 5. This means that 9B has 5 electrons, 5 protons and 4 neutrons. For 11B, the isotope has 5 protons, 5 electrons and 6 neutrons. 9 and 11 are the mass number of the isotopes; mass number is the summation of number of proton and neutron.
5 0
3 years ago
Can someone help please?
dedylja [7]
Hey are you from Calvert too? I got the same questions for the DBA
8 0
3 years ago
The decomposition of nitrogen dioxide to nitrogen monoxide and oxygen gas is a second order process as suspected in the previous
lesya [120]

Explanation:

For the given reaction 2NO_{2} \rightarrow 2NO + O_{2}

Now, expression for half-life of a second order reaction is as follows.

                  t_{1/2} = \frac{1}{[A_{0}]k}     ....... (1)

Second half life of this reaction will be t_{1/4}. So, expression for this will be as follows.

          t_{1/4} = \frac{1}{k} [\frac{1}{[A]_{f}} - \frac{1}{[A_{0}]}]  ...(2)

where [A]_{f} is the final concentration that is, \frac{[A]_{0}}{4} here and [A]_{i} is the initial concentration.

Hence, putting these values into equation (2) formula as follows.

        t_{1/4} = \frac{1}{k} [\frac{4}{[A]_{0}} - \frac{1}{[A_{0}]}]

                      = \frac{3}{[A_{0}]k}     ...... (3)

Now, dividing equation (3) by equation (1) as follows.

           \frac{t_{1/4}}{t_{1/2}} = \frac{3}{[A_{0}]k} \times [A_{0}]k                  

                                    = 3

or,                       t_{1/4} = 3 t_{1/2}    

Thus, we can conclude that one would expect the second half-life of this reaction to be three times the first half-life of this reaction.

6 0
3 years ago
When the glucose solution described in part A is connected to an unknown solution via a semipermeable membrane, the unknown solu
algol13

Answer:

B

Explanation:

4 0
3 years ago
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