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3 years ago

5 drops of 0.15 M Ki added to40 drops of Na2S2O3What is the final concentration of ki?

Chemistry

1 answer:

rjkz [21]3 years ago

3 0

Answer:

$\boxed{\text{0.017 mol/L}}$

Explanation:

Na₂S₂O₃ solution does not react with KI (it reacts with I₂), so it is simply diluting the KI, and we can use the dilution formula.

$c_{1}V_{1} = c_{2}V_{2}$

Data:

c₁ = 0.15 mol·L⁻¹; V₁ = 5 drops

V(Na₂S₂O₃) = 40 drops

Calculations:

(a) Calculate the total volume

V₂ = 5 + 40 = 45 drops

(b) Calculate the concentration

0.15 × 5 = c₂ × 45

0.75 = 45c₂

$c_{2} = \dfrac{0.75 }{45} = \boxed{\textbf{0.017 mol/L}}$

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2 years ago

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3 years ago

.500 mol of br2 and .500 mol of cl2 are placed in a .500L flask and allowed to reach equilibrium. at equilibrium the flask was f

Tasya [4]

Answer : The value of $K_c$ for the given reaction is, 0.36

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

$Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)$

The expression of $K_c$ will be,

$K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}$

First we have to calculate the concentration of $Br_2,Cl_2\text{ and }BrCl$ .

$\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M$

$\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M$

$\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M$

Now we have to calculate the value of $K_c$ for the given reaction.

$K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}$

$K_c=\frac{(0.6)^2}{(1)\times (1)}$

$K_c=0.36$

Therefore, the value of $K_c$ for the given reaction is, 0.36

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3 years ago

Read 2 more answers

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chubhunter [2.5K]

Answer:

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Explanation:

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3 years ago

What pressure will be exerted by 25 g of CO2 at temperature of 25°C and a volume of .50 L?

Grace [21]

Answer:

P = 27.9 atm

Explanation:

Given data:

Mass of CO₂ = 25 g

Temperature = 25°C (25+273.15 K = 298.15 K)

Volume of gas = 0.50 L

Pressure of gas = ?

Solution:

Firs of all we will calculate the number of moles of gas,

Number of moles = mass/molar mass

Number of moles = 25 g/ 44 g/mol

Number of moles = 0.57 mol

Pressure of gas :

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

P × 0.50 L = 0.57 mol × 0.0821 atm.L/ mol.K × 298.15 K

P = 13.95 atm.L/ 0.50 L

P = 27.9 atm

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3 years ago

5 drops of 0.15 M Ki added to40 drops of Na2S2O3What is the final concentration of ki?​

5 drops of 0.15 M Ki added to40 drops of Na2S2O3What is the final concentration of ki?