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JulsSmile [24]
2 years ago
10

Under standard conditions, a given reaction is endergonic (i.e., ΔG >0). Which of the following can render this reaction favo

rable: using the product immediately in the next step, maintaining a high starting-material concentration, or keeping a high product concentration? View Available Hint(s) Under standard conditions, a given reaction is endergonic (i.e., ΔG >0). Which of the following can render this reaction favorable: using the product immediately in the next step, maintaining a high starting-material concentration, or keeping a high product concentration? maintaining a high starting-material concentration keeping a high product concentration using the product immediately in the next step and maintaining a high starting-material concentration using the product immediately in the next step
Chemistry
1 answer:
sleet_krkn [62]2 years ago
5 0

Answer:

Maintaining a high starting-material concentration can render this reaction favorable.

Explanation:

A reaction is <em>favorable</em> when <em>ΔG < 0</em> (<em>exergonic</em>). ΔG depends on the temperature and on the reaction of reactants and products as established in 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

Q is the reaction quotient

To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.

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Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic
Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = 164.5\times 10^{-9} g

1 ng=10^{-9} g

Volume of the sample = V = 15.3 cm^3

Density of the lake water sample ,d= 1.00 g/cm^3

Mass of sample =  M = d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g

ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in 1 cm^3  of lake water = \frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g

Mass of arsenic in 0.710 km^3 lake water be m.

1 km^3=10^{15} cm^3

Mass of arsenic in 0.710\times 10^{15} cm^3 lake water :

m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

It will take 1.37 years to remove all of the arsenic from the lake.

3 0
2 years ago
At elevated temperatures, methylisonitrile (CH3NC) isomerizes to acetonitrile (CH3CN): CH3NC (g) → CH3CN (g) At the start of the
Olegator [25]

Answer:

Average rate of decomposition = 3.68 x 10⁻³ mol/min

Explanation:

CH₃NC (g) → CH₃CN (g)

Average rate of decomposition = [CH₃NC]/Δt

[CH₃NC] = initial [CH₃NC] - final [CH₃NC]

[CH₃NC] = 0.200 mol - 0.108 mol

[CH₃NC] = 0.092 mol

Average rate of decomposition = 0.092 mol / 25 min

Average rate of decomposition = 3.68 x 10⁻³ mol/min

7 0
2 years ago
Calculate the heat change in calories for melting 65 g of ice at 0 ∘c.
Genrish500 [490]

When ice melts, the physicals state changes from solid to liquid. The energy or the heat required (q) required to change a unit mass (m) of a substance from solid to liquid is known as the enthalpy or heat of fusion (ΔHf). The variables; q, m and ΔHf are related as:

q = m * ΔHf

the mass of ice m = 65 g

the heat of fusion of water at 0C = ΔHf = 334 J/g

Therefore: q = 65 g * 334 J/g = 21710 J

Now:

4.184 J = 1 cal

which implies that: 21710 J = 1 cal * 21710 J/4.184 J = 5188.8 cal

Hence the heat required is 5188.8 cal or 5.2 Kcal (approx)

5 0
2 years ago
A 20.0 -\,L volume of an ideal gas in a cylinder with a piston is at a pressure of 3.2 atm. Enough weight is suddenly removed fr
zzz [600]

Answer:

1. ΔE = 0 J

2. ΔH = 0 J

3. q = 3.2 × 10³ J

4. w = -3.2 × 10³ J

Explanation:

The change in the internal energy (ΔE) and the change in the enthalpy (ΔH) are functions of the temperature. If the temperature is constant, ΔE = 0 and ΔH = 0.

The gas initially occupies a volume V₁ = 20.0 L at P₁ = 3.2 atm. When the pressure changes to P₂ = 1.6 atm, we can find the volume V₂ using Boyle's law.

P₁ × V₁ = P₂ × V₂

3.2 atm × 20.0 L = 1.6 atm × V₂

V₂ = 40 L

The work (w) can be calculated using the following expression.

w = - P . ΔV

where,

P is the external pressure for which the process happened

ΔV is the change in the volume

w = -1.6 atm × (40L - 20.0L) = -32 atm.L × (101.325 J/1atm.L) = -3.2 × 10³ J

The change in the internal energy is:

ΔE = q + w

0 = q + w

q = - w = 3.2 × 10³ J

6 0
2 years ago
What is the net ionic equation of the reaction of MgCl2 with NaOH? Express your answer as a chemical equation. View Available Hi
kari74 [83]

Answer:

Net ionic equation for the reaction between MgCl₂ and NaOH in water:

\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s).

Net ionic equation for the reaction between MgSO₄ and BaCl₂ in water:

\rm {Ba}^{2+}\; (aq) + {SO_4}^{2-}\;(aq) \to BaSO_4\; (s).

Explanation:

Start by finding the chemical equations for each reaction:

MgCl₂ reacts with NaOH to form Mg(OH)₂ and NaCl. This reaction is a double decomposition reaction (a.k.a. double replacement reaction, salt metathesis reaction.) This reaction is feasible because one of the products, Mg(OH)₂, is weakly soluble in water and exists as a solid precipitate.

\rm MgCl_2\; (aq) + 2\; NaOH\; (aq)\to Mg(OH)_2 \; (s) + 2\; NaCl\; (aq).

MgSO₄ reacts with BaCl₂ in a double decomposition reaction to produce BaSO₄ and MgCl₂. Similarly, the solid product BaSO₄ makes this reaction is feasible.

\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq).

How to rewrite a chemical equation to produce a net ionic equation?

  1. Rewrite all reactants and products that ionizes completely in the solution as ions.
  2. Eliminate ions that exist on both sides of the equation to produce a net ionic equation.

Typical classes of chemicals that ionize completely in water:

  • Soluble salts,
  • Strong acids, and
  • Strong bases.

Keep the formula of salts that are not soluble in water, weak acids, weak bases, and water unchanged.

Take the first reaction as an example, note the coefficients:

  • MgCl₂ is a salt and is soluble in water. Each unit of MgCl₂ can be written as \rm Mg^{2+} and \rm 2\; Cl^{-}.
  • NaOH is a strong base. Each unit of NaOH can be written as \rm Na^{+} and \rm OH^{-}.
  • Mg(OH)₂ is a weak base and should not be written.
  • NaCl is a salt and is soluble in water. Each unit of NaCl can be written as \rm Na^{+} and \rm Cl^{-}.

\rm Mg^{2+} + 2\; Cl^{-} + 2\; Na^{+} + 2\; OH^{-} \to Mg(OH)_2\;(s) + 2\; Na^{+} + 2\; Cl^{-}.

Ions on both sides of the equation:

  • \rm 2\; Cl^{-}, and
  • \rm 2\; Na^{+}.

Add the state symbols:

\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s).

For the second reaction:

\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq).

\rm Mg^{2+} + 2\; {SO_4}^{2-} + Ba^{2+} + 2\; Cl^{-} \to BaSO_4\; (s) + Mg^{2+} + 2\; Cl^{-}.

\rm Ba^{2+}\; (aq) + {SO_4}^{2-}\; (aq) \to BaSO_4\; (s).

5 0
2 years ago
Read 2 more answers
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