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3 years ago

5

The Earth’s circumference around the equator is slightly bulged as compared to the circumference around the poles. If the circum

ference around Earth’s equator is 29,902 miles, what is this in Mm?

1 answer:

Elanso [62]3 years ago

8 0

Answer: 48.123 Mm

Explanation:

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What information can a foliated metamorphic rock provide you about the conditions under which it formed?

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A foliated rock forms when the minerals are realigned due to the presence of high temperature and pressure. The minerals align in such a way that the axes are directed to where the pressure was applied.

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All of the following are examples of matter except

Alexxandr [17]

A. Heat because it does not take up space.

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3 years ago

How many 325mg aspirin tablets can be made from 3628 grams of aspirins ? Need an answer by tonight it’s getting late .

krek1111 [17]

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4 years ago

Calculate the activity coefficients for the following conditions:

uysha [10]

<u>Answer:</u>

<u>For a:</u> The activity coefficient of copper ions is 0.676

<u>For b:</u> The activity coefficient of potassium ions is 0.851

<u>For c:</u> The activity coefficient of potassium ions is 0.794

<u>Explanation:</u>

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

<u>For a:</u>

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

<u>For b:</u>

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

<u>For c:</u>

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

The activation energy of an uncatalyzed reaction is 95 kJ / mol. The addition of a catalyst lowers the activation energy to 55 k

dybincka [34]

Answer:

The rate of the catalyzed reaction will increase by a 1.8 × 10⁵ factor.

Explanation:

The rate of a reaction (r) is proportional to the rate constant (k). We can find the rate constant using the Arrhenius equation.

$k=A.e^{-Ea/R.T}$

where,

A: collision factor

Ea: activation energy

R: ideal gas constant

T: absolute temperature (125°C + 273 = 398 K)

For the uncatalized reaction,

$kU=A.e^{-95\times 10^{3} kJ/mol /(8.314J/K.mol).398K}=3.4\times 10^{-13}A$

For the catalized reaction,

$kC=A.e^{-55\times 10^{3} kJ/mol /(8.314J/K.mol).398K}=6.0\times 10^{-8}A$

The ratio kC to kU is 6.0 × 10⁻⁸A/3.4 × 10⁻¹³A = 1.8 × 10⁵

8 0

4 years ago