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olga_2 [115]
3 years ago
15

Use your periodic table to answer the following questions. Pay attention to the spelling of the element names and make sure to w

rite the chemical symbols correctly. 1. What metal is found in the group 1 and period 4? __________________ 2. What metalloid is found in the group 16 and period 5? ______________ 3. What nonmetal is found in the group 17 and period 2? ______________ 4. In what group will you find copper? ____________________ 5. In what group will you find krypton? ____________________ 6. In what group will you find mercury? _______________________ 7. What is the symbol for chlorine? ___________________ 8. What is the symbol for lead? _____________________ 9. What is the symbol for tungsten? ______________________ 10. What is the symbol for antimony? _______________________ 11. What is the symbol for sodium? ______________________ 12. In what period will you find gold? __________________________ 13. In what period will you find silver? _________________________ 14. In what period will you find oxygen? _________________________ 15. In what groups will you find metalloids? ______ ______ _____ _____ 16. What element has the symbol H? _______________________ 17. What element has the symbol He? _______________________ 18. What element has the symbol Ca? ________________________ 19. What element has the symbol Co? ________________________ 20. What element has the symbol C?
Chemistry
1 answer:
Novay_Z [31]3 years ago
3 0

Explanation:

1. K (Potassium)

2. Te (Tellurium)

3. F (Flourine)

4. group 11, period 4

5. group 18, period 4

6. group 12, period 6

7. Cl

8. Pb

9. W

10. Sb

11. Na

12. period 6, group 11

13. period 5, group 11

14. period 2, group 16

15. Groups 13–16 of the periodic table contain one or more metalloids

16. Hydrogen

17. Helium

18. Calcium

19. Cobalt

20. Carbon

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carbondioxide is an example of compound.

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Can somebody plz help answer all these true and false questions CORRECT!!
goldfiish [28.3K]

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7 0
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Which of the following would have the largest amount of mass?
Elodia [21]

Answer:

1 kg

Explanation:

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kilo 10^3

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7 0
3 years ago
Read 2 more answers
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
You are given a sulfuric acid solution of unknown concentration. You dispense 10.00 mL of the unknown solution into an Erlenmeye
dmitriy555 [2]

Answer:

"0.053457 M" of sulfuric acid.

Explanation:

The given values are:

V = 10 mL solution

V_{added} = 12.20 mL

V_{total} = 22.20 mL

then,

M 0.103 M of NaOH,

V_{rinsed} = experiment  will not be affected

V_{total \ base} = 10.38 mL

Now,

⇒  mol of NAOH = MV

                            = 0.103\times 10.38

                            =  1.06914  \ m

Whether Sulfuric acid, then

⇒  H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O

⇒  mol \ of \ acid =\frac{1}{2}\times \ mol \   of  \ base

⇒  1.06914 \ m \ mol \ of \ base = \frac{1}{2}\times 1.06914 = 0.53457 \ m \ mol \ of \ acid

Before any dilution:

V_{sample} = 10  \ mL

⇒  M \ acid = \frac{m \ mol}{V}

                 =\frac{ 0.53457 }{10}

                 =0.053457 \ M (Sulfuric acid)

6 0
3 years ago
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