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victus00 [196]
1 year ago
11

B-ad=3c+r solve for c

Chemistry
1 answer:
DENIUS [597]1 year ago
7 0

By making c the subject of formula in the given equation, the value "c" is equal to c = (B - ad - r)/3.

<h3>What is an equation?</h3>

An equation can be defined as a mathematical expression which shows that two (2) or more thing are equal.

In this exercise, you're required to solve for c from the given mathematical expression (equation) by making it the subject of formula. This ultimately implies that, all the other variables would be defined in terms of c and they would all be on the same side of the "equal to" symbol.

Making c the subject of formula, we have:

B - ad = 3c + r

3c = B - ad - r

Dividing both sides by 3, we have:

3c/3 = (B - ad - r)/3

c = (B - ad - r)/3.

Read more on subject of formula here: brainly.com/question/21140562

#SPJ1

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Answer:

chlorine is therefore an anion

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3 years ago
In general, in what type of solvent (non-polar, moderately polar, or highly polar) are polar solutes most soluble? Explain why.
tia_tia [17]

Answer:

  • In general, polar solutes are most soluble in highly polar solvents.

Explanation:

The general rule is "like dissolves like" which means that <em>polar solvents </em>dissolve polar (or ionic) <em>solutes</em> and <em>non-polar solvents</em> dissolve non-polar solutes.

In order for a solvent dissolve a solute, the strength of the interacttion (force) between the solute and the solvent units (atoms, molecules, or ions) must be stronger than the strength of the forces that keep together he particles of the pure substances (known as intermolecular forces).

Since the nature of the interactions between the units are electrostatic, the more polar is the solvent the better it will be able to attract and surround the solute particles, keeping them separated and in solution. That mechanism explains why polar solutes will be most soluble in highly polar solvents.

5 0
3 years ago
HELP PLEASE DUE SOON!!!
marishachu [46]

Answer: 129.93577

Explanation:

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6 0
3 years ago
Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi
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Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g)     ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

[]=\frac{P}{R.T}

<em>Calculate ΔG at 25°C given the following sets of partial pressures.</em>

<em>Part A  130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.</em>

[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M

[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

<em>Part B  5.0atm SO₂, 3.0atm O₂, 30atm SO₃  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M

[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M

[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

<em>Part C Each reactant and product at a partial pressure of 1.0 atm.  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

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3 years ago
The diagram shows the thermohaline circulation, also known as the ocean's
svetoff [14.1K]

Answer:

A. Water in the blue currents is colder, saltier, and denser than water

in the red currents.

Explanation:

a p e x

3 0
2 years ago
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