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3 years ago

Help me please? I need to pass this tomorrow

Chemistry

1 answer:

kramer3 years ago

6 0

Here is what I have come up with, and I have used Mariam Webster for the definitions.

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4 0

3 years ago

Read 2 more answers

State Modern periodic Table

Kaylis [27]

Answer:

According to the modern periodic law, the properties of the elements and their compounds are a periodic function of their atomic numbers. Thus, in the modern periodic table, atomic number forms the basis of the classification of elements;The modern table is called 'long form' of the periodic table.

hope this helps

5 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

(I)how many atoms are present in 7g of lithium?

ICE Princess25 [194]

Answer :

(i) The number of atoms present in 7 g of lithium are, $6.07\times 10^{23}$

(ii) The number of atoms present in 7 g of lithium are, $1.204\times 10^{24}$

(iii) The number of moles of $F_2$ is, 1 mole

The number of moles of $CO_2$ is, 0.5 mole

The number of moles of $OH^-$ is, 1 mole

Explanation :

Part (i) :

First we have to calculate the moles of lithium.

$\text{Moles of }Li=\frac{\text{Mass of }Li}{\text{Molar mass of }Li}$

Molar mass of Li = 6.94 g/mole

$\text{Moles of }Li=\frac{7g}{6.94g/mol}=1.008mole$

Now we have to calculate the number of atoms present.

As, 1 mole of lithium contains $6.022\times 10^{23}$ number of atoms

So, 1.008 mole of lithium contains $1.008\times 6.022\times 10^{23}=6.07\times 10^{23}$ number of atoms

Thus, the number of atoms present in 7 g of lithium are, $6.07\times 10^{23}$

Part (ii) :

First we have to calculate the moles of carbon.

$\text{Moles of }C=\frac{\text{Mass of }C}{\text{Molar mass of }C}$

Molar mass of C = 12 g/mole

$\text{Moles of }C=\frac{24g}{12g/mol}=2mole$

Now we have to calculate the number of atoms present.

As, 1 mole of carbon contains $6.022\times 10^{23}$ number of atoms

So, 2 mole of carbon contains $2\times 6.022\times 10^{23}=1.204\times 10^{24}$ number of atoms

Thus, the number of atoms present in 7 g of lithium are, $1.204\times 10^{24}$

Part (iii) :

To calculate the moles of $F_2$ :

$\text{Moles of }F_2=\frac{\text{Mass of }F_2}{\text{Molar mass of }F_2}$

Molar mass of $F_2$ = 38 g/mole

$\text{Moles of }F_2=\frac{19g}{19g/mol}=1mole$

Thus, the number of moles of $F_2$ is, 1 mole

To calculate the moles of $CO_2$ :

$\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}$

Molar mass of $CO_2$ = 44 g/mole

$\text{Moles of }CO_2=\frac{22g}{44g/mol}=0.5mole$

Thus, the number of moles of $CO_2$ is, 0.5 mole

To calculate the moles of $OH^-$ ions :

$\text{Moles of }OH^-=\frac{\text{Mass of }OH^-}{\text{Molar mass of }OH^-}$

Molar mass of $OH^-$ = 17 g/mole

$\text{Moles of }OH^-=\frac{17g}{17g/mol}=1mole$

Thus, the number of moles of $OH^-$ is, 1 mole

4 0

3 years ago

Which color in the visible spectrum of hydrogen has the least energy

astraxan [27]

I believe it would be the color Red.

4 0

3 years ago

Help me please? I need to pass this tomorrow ​

Help me please? I need to pass this tomorrow