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3 years ago

This is a personal question Should I tryout for drumline??

Chemistry

2 answers:

matrenka [14]3 years ago

4 0

Uhm, of course you should! Have fun!

dezoksy [38]3 years ago

3 0

Answer:

Yes

Explanation:

Periodt

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A hypothetical wave has 4.6 J of energy. What is its hypothetical, approximate frequency?

Mariulka [41]

Answer:

6.94 × 10^33Hz

Explanation:

E = hf

Where;

E = Energy of wave (J)

h = Planck's constant (6.626 × 10^-34J/s)

f = frequency (Hz)

According to the information provided in this question, the hypothetical energy of the wave is 4.6J

Hence, using E = hf

4.6 = 6.626 × 10^-34 × f

f = 4.6 / 6.626 × 10^-34

f = 0.694 × 10^34

f = 6.94 × 10^33Hz

3 0

3 years ago

Please help I need I badly or I'm going to fail

Nimfa-mama [501]

Answer:

Its B

Explanation:

4 0

3 years ago

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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

Measurements show that unknown compound has the following composition: element mass 62.1 % carbon, 10.5 % hydrogen and 27.6 % ox

tino4ka555 [31]

Answer:

C3H6O

Explanation:

The percentage composition of the elements in the compound are given as follows:

62.1 % carbon = 62.1g of C

10.5 % hydrogen = 10.5g of H

27.6 % oxygen = 27.6g of O

Next, we convert each mass to mole by dividing by their molar/atomic mass

C = 62.1/12 = 5.175mol

H = 10.5/1 = 10.5mol

O = 27.6/16 = 1.725mol

Next, we divide each mole value by the smallest mole value (1.725)

C = 5.175mol ÷ 1.725 = 3

H = 10.5mol ÷ 1.725 = 6.086

O = 1.725mol ÷ 1.725 = 1

The empirical ratio approximately of C:H:O is 3:6:1, hence, the empirical formula is C3H6O

5 0

3 years ago

Hi can you help me with these questions please

joja [24]

Answer:

iron oxide11

Explanation:

iron oxide is the chemical compound with formula fe304. it occurs in nature as the mineral magnetite.

6 0

3 years ago

Read 2 more answers