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rewona [7]
3 years ago
13

The wastewater solution from a factory containing high levels of salts needs to be diluted before it can be released into the en

vironment. Therefore, two containers of waste solution are separated by a semipermeable membrane and pressure is applied to one container, forcing only water molecules through the membrane and diluting the waste solution in the other container. As dilution continues, higher and higher pressures are needed to counteract the natural tendency for the water molecules to have a net flow back toward the more concentrated solution.
What was the applied pressure at the end of this process if the final concentrations of the solutions were 0.048 M and 0.190 M at a temperature of 23 C?
Chemistry
1 answer:
Vesna [10]3 years ago
3 0

Answer:

3.5 atm

Explanation:

As stated in the question pressure is required to counteract the natural tendency  for water to dilute the more concentrated solution. The difference in concentrations will give us the answer using the osmotic pressure equation.

π = ( n/v)  RT where n/v is the molarity (mol/L), R is the gas constant and T is the temperature.

The difference in osmotic pressure of the solutions is:

Δπ = Δ c RT where c is the difference in molar concentrations.

pressure required = Δπ = (0.190 - 0.048) M x 0.821 Latm/Kmol x 298 K

= 3.47 atm

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What is the coefficient of Ca(OH)2 in the equation Ca(OH)2 + HNO3 → Ca(NO3)2 + H2O when balanced using the smallest possible coe
Vikentia [17]

Answer:

The coefficient of Ca(OH)2 is 1

Explanation:

Step 1: unbalanced equation

Ca(OH)2 + HNO3 → Ca(NO3)2 + H2O

Step 2: Balancing the equation

On the right side we have 2x N (in Ca(NO3)2 ) and 1x N on the left side (in HNO3). To balance the amount of N on both sides, we have to multiply HNO3 by 2.

Ca(OH)2 + 2HNO3 → Ca(NO3)2 + H2O

On the left side we have 4x H (2xH in Ca(OH)2 and 2x H in HNO3), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side, by 2.

Now the equationis balanced.

Ca(OH)2 + 2HNO3 = Ca(NO3)2 + 2H2O

The coefficient of Ca(OH)2 is 1

4 0
3 years ago
Potassium metal and chlorine gas combine in a reaction to produce an ionic compound. What is the correct balanced equation for t
Alisiya [41]

Answer:

2K (s) + Cl₂ (g) ⇒ 2KCl (s)

Explanation:

Potassium and chlorine gas combine to form potassium chloride which is an ionic compound. The reaction is a type of combination reaction in which chlorine is being added to the metal, potassium.

Potassium reacts violently with the chlorine which is yellowish green in color to produce white solid of potassium chloride.

The balanced reaction is shown below as:

2K (s) + Cl₂ (g) ⇒ 2KCl (s)

6 0
3 years ago
If the freezing point of the solution had been incorrectly read 0.3 °C lower than the true freezing point, would the calculated
Dovator [93]

Answer : The molar mass of the solute would be low.

Explanation :

Formula used for depression in freezing point is:  

\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{w_b}{M_b}\times w_a}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution

\Delta T^o = freezing point of water

i = Van't Hoff factor

K_f = freezing point constant

m = molality

w_b = mass of solute

w_a = mass of solvent

M_b = molar mass of solute

From the formula we conclude that, when the freezing point of the solution read incorrectly that is freezing point of the solution is lower than the true freezing point then this means that change in freezing point would be high and the molar mass of the solute would be low.

Hence, the molar mass of the solute would be low.

6 0
3 years ago
ASAP PLS
AveGali [126]

Answer:

think I did this before and its V

8 0
3 years ago
Suppose 6.63g of zinc bromide is dissolved in 100.mL of a 0.60 M aqueous solution of potassium carbonate. Calculate the final mo
Alenkasestr [34]

Answer:

[Zn²⁺] = 4.78x10⁻¹⁰M

Explanation:

Based on the reaction:

ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)

The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:

1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]

We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:

<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>

6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺

<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>

0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻

Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =

0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]

Replacing in Ksp expression:

1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]

<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>

4 0
3 years ago
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