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2 years ago

Find the percentage compositions CaCl2 Ca: CI:

Chemistry

1 answer:

Naddika [18.5K]2 years ago

4 0

Answer:

PERCENTAGE COMPOSITION OF CALCIUM IS 40.1/74.1

PERCENTAGE OF CHLORINE IS 17/74.1

Explanation:

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Reaction rate is increased by an increase in kinetic energy

marta [7]

Answer:

Reaction rate is increased by an increase in kinetic energy

of the reacting particles when their

is _temperature_

increased

A . Collision theory

B . Surface Area

C . Temperature

D. Concentration

8 0

2 years ago

A diver has a lung capacity of 2.4 L when the pressure is 0.8 atm. What is the volume of the diver’s lungs when the pressure cha

tangare [24]

Answer:

The final volume is 1.6 L.

Explanation:

It is given that,

A diver has a lung capacity of 2.4 L when the pressure is 0.8 atm. We need to find the volume of the diver’s lungs when the pressure changes to 1.2 atm. Let V₂ is volume.

It is based on Boyle's law. According to this law,

$PV=K$

K is constant

$P_1V_1=P_2V_2$

$V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{0.8\times 2.4}{1.2}\\\\V_2=1.6\ L$

So, the final volume is 1.6 L.

7 0

3 years ago

On the periodic table, in which period is copper?

Firdavs [7]

Answer:

At the top of Group 11 above silver and gold.

Period 4

Explanation:

3 0

2 years ago

Given 7.45 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100

KengaRu [80]

The equation for the reaction is:

C₄H₈O₂ + C₂H₅OH = C₆H₁₂O₂ + H₂O

Now you see that the number of the moles of butanoic acid and etyl butyrate is equal in

the reaction. That means;

number of moles of C₄H₈O₂ = number of moles of C₆H₁₂O₂

mass of C₄H₈O₂/ Molar mass of C₄H₈O₂ = mass of C₆H₁₂O₂/ molar mass of C₆H₁₂O₂

mass of C₆H₁₂O₂ = molar mass of C₆H₁₂O₂ x mass of C₄H₈O₂/ Molar mass of C₄H₈O₂

Now, assuming 100% yield, the mass of ethyl butyrate produced is:

= 7.45/88.11 x 116.16

=9.82g

Thus, the theoretical yield of ethyl butyrate is 9.82g.

3 0

3 years ago

Calculate the activity coefficients for the following conditions:

uysha [10]

Answer:

For a: The activity coefficient of copper ions is 0.676

For b: The activity coefficient of potassium ions is 0.851

For c: The activity coefficient of potassium ions is 0.794

Explanation:

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

$-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}$ ........(1)

where,

$\gamma_i$ = activity coefficient of ion

$Z_i$ = charge of the ion

$\mu$ = ionic strength of solution

$\alpha _i$ = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

$\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)$ ......(2)

where,

$C_i$ = concentration of i-th ions

$Z_i$ = charge of i-th ions

For a:

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

$C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M$

Now, calculating the activity coefficient of $Cu^{2+}$ ion in the solution by using equation 1:

$Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{ (known)}\\\mu=0.01M$

Putting values in equation 1, we get:

$-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676$

Hence, the activity coefficient of copper ions is 0.676

For b:

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

$C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.025M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851$

Hence, the activity coefficient of potassium ions is 0.851

For c:

We are given:

0.02 M $K_2SO_4$ solution:

Calculating the ionic strength by using equation 2:

$C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.06M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794$

Hence, the activity coefficient of potassium ions is 0.794

6 0

3 years ago