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3 years ago

What volume of water is needed for 0.196 mol of CdCl2 to prepare a 3.91 M solution? L

Chemistry

1 answer:

Brilliant_brown [7]3 years ago

4 0

0.050 litres of the water will be required to make a 3.91 M solution with 0.196 moles in it.

Explanation:

Data given:

moles of Cd $Cl_{2}$ = 0.196 Moles

Molarity of the solution = 3.91 M

Volume in litres =?

molarity is calculated by the formula:

molarity = $\frac{number of moles}{volume in litres}$

putting the values in the above formula and rearranging it for volume:

volume = $\frac{number of moles}{molarity}$

volume = $\frac{0.196}{3.91}$

volume = 0.050 litres

0.050 litres of the water will be required to make a 3.91 M solution with 0.196 moles in it.

Molarity is the number of moles present in a given volume of solution which is given in litres. It is the measurement of the concentration of particular solute in a solution.

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Nitrogen has two isotopes. One has an atomic mass of 14.003

Ilia_Sergeevich [38]

Answer:

d= 14.007 amu

Explanation:

Abundance of N¹⁴ = 99.63%

Abundance of N¹⁵ = 0.37%

Atomic mass of N¹⁴ = 14.003 amu

Atomic mass of N¹⁵ = 15.000 amu

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100

Average atomic mass = (14.003 × 99.63)+(15.000× 0.37) /100

Average atomic mass = 1395.12 + 5.55 / 100

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3 years ago

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

Now put all the given values in the above formula, we get:

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

Now put all the given values in the above formula, we get:

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

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3 years ago

What is the electron configuration of Zn2+

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. 1s2 2s2 2p6 3s2 3p6 3d10 One point is earned for the correct configuration.

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Select each process that shows a physical change. Select all that apply.

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Answer:

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The reaction between solid white phosphorous and oxygen produces solid tetraphosphorous decaoxide (P4010).

Alchen [17]

Answer:

oxygen is limiting reactant

Explanation:

Given data:

Mass of phosphorus = 25.0 g

Mass of oxygen = 50.0 g

What is limiting reactant ?

Solution:

Chemical equation:

P₄ + 5O₂ → P₄O₁₀

Number of moles of P₄:

Number of moles = mass/molar mass

Number of moles = 25.0 g/ 123.89 g/mol

Number of moles = 0.20 mol

Number of moles of O₂:

Number of moles = mass/molar mass

Number of moles = 50.0 g/ 32 g/mol

Number of moles = 1.56 mol

now we will compare the moles of reactants with product:

P₄ : P₄O₁₀

1 : 1

0.20 : 0.20

O₂ : P₄O₁₀

5 : 1

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Less number of moles of product are formed by the oxygen thus it will act as limiting reactant.

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3 years ago