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denis23 [38]
2 years ago
5

To make a solution of lemonade, lemon juice and sugar are dissolved in water. the water is the ______ in this example.

Chemistry
1 answer:
vova2212 [387]2 years ago
8 0

To make a solution of lemonade, lemon juice and sugar are dissolved in water. the water is the <u>Solvent </u>in this example.

Solvent Examples. Common examples of solvents consist of water, ethanol, methanol, and acetone. The term 'solvent' may be defined as a substance that has the ability to dissolve a given solute to form a solution with it.

Solvents are substances that are used to dissolve the solutes used in the formulation. These solutes can be solids, liquids, or gaseous in nature. Thus solvents are used to get a solution upon interacting the solute with a suitable solvent.

Solvents are usually, however not usually, liquids. They can also be gases or solids. The material dissolved in the solvent is referred to as the solute. collectively, the solvent and solute comprise the solution.

Learn more about Solvent here brainly.com/question/25326161

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Kepler's Law of Planetary Motion __________ the movement of planets in the universe
ipn [44]
Kepler's Law of Planetary Motion consists the movement of planets in the universe

if not? what are the four or three answers for me to answer?
3 0
3 years ago
Using the balanced equation below,
Zielflug [23.3K]

Answer:

Mass = 6.538 g

Explanation:

Given data:

Mass of zinc hydroxide produced = 9.65 g

Mass of zinc required = ?

Solution:

Chemical equation:

Zn + 2MnO₂ + H₂O      →      Zn(OH)₂ + Mn₂O₃

Number of moles of zinc hydroxide:

Number of moles = mass/molar mass

Number of moles = 9.65 g/ 99.42 g/mol

Number of moles = 0.1 mol

now we will compare the moles of zinc and zinc hydroxide,

                  Zn(OH)₂           :          Zn

                       1                  :          1

                    0.1                  :         0.1

Mass of zinc required:

Mass = number of moles × molar mass

Mass = 0.1 mol × 65.38 g/mol

Mass = 6.538 g

8 0
3 years ago
Read 2 more answers
Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms
nirvana33 [79]

Answer:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }

With further simplification, we have:

P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]

Then, we have:

P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]

Therefore,

PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]

Using the expansion:

(1-x)^{-1} = 1 + x + x^{2} + ....

Therefore,

PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]

Thus:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]           equation (1)

Using the virial equation of state:

P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]

Thus:

PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]     equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol[/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0
3 years ago
Balance the following skeleton reactions and identify the oxidizing and reducing agents:(b) P₄(s) → HPO₃²⁻(aq) + PH₃(g) [acidic]
m_a_m_a [10]

The balanced chemical equation is :

5P₄ + 36OH → 12HPO₃⁻² (aq) + 8PH₃ (acidic)

Here the oxidation number of P changed from 0 to -3 in PH₃ and increases  from 0 to +3 in HPO₃⁻². When P₄ changes to PH₃ reduction reaction is taking place as there is addition of hydrogen and when P₄ changes to HPO₃⁻² oxidation takes place as there is addition of oxygen.

Thus clearly both reduction and oxidation are taking place.

Thus, we can infer that here P₄ is both oxidizing as well as reducing agent.

To know more about oxidation number here:

brainly.com/question/13182308

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5 0
2 years ago
Water molecules are highly _______ and are always _________.
anyanavicka [17]

Water molecules are highly packed and are always near each other.

7 0
3 years ago
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