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anygoal [31]
2 years ago
13

Ionic bonds are formed by the attraction between two ________________.

Chemistry
1 answer:
cricket20 [7]2 years ago
7 0
It would be C) atoms of the opposite charges. iconic bonds are formed by the attraction of two atoms of the opposite charge.

Hope this helps :)

can you please make this the brainiest answer it would really help me. thank you :) 
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6 0
2 years ago
The rock in a lead ore deposit contains 89 % PbS by mass. How many kilograms of the rock must be processed to obtain 1.5 kg of P
Zolol [24]

Answer:

Approximately 1.9 kilograms of this rock.

Explanation:

Relative atomic mass data from a modern periodic table:

  • Pb: 207.2;
  • S: 32.06.

To answer this question, start by finding the mass of Pb in each kilogram of this rock.

89% of the rock is \rm PbS. There will be 890 grams of \rm PbS in one kilogram of this rock.

Formula mass of \rm PbS:

M(\mathrm{PbS}) = 207.2 + 32.06 = 239.26\; g\cdot mol^{-1}.

How many moles of \rm PbS formula units in that 890 grams of \rm PbS?

\displaystyle n = \frac{m}{M} = \rm \frac{890}{239.26} = 3.71980\; mol.

There's one mole of \rm Pb in each mole of \rm PbS. There are thus \rm 3.71980\; mol of \rm Pb in one kilogram of this rock.

What will be the mass of that \rm 3.71980\; mol of \rm Pb?

m(\mathrm{Pb}) = n(\mathrm{Pb}) \cdot M(\mathrm{Pb}) = \rm 3.71980 \times 207.2 = 770.743\; g = 0.770743\; kg.

In other words, the \rm PbS in 1 kilogram of this rock contains \rm 0.770743\; kg of lead \rm Pb.

How many kilograms of the rock will contain enough \rm PbS to provide 1.5 kilogram of \rm Pb?

\displaystyle \frac{1.5}{0.770743} \approx \rm 1.9\; kg.

5 0
3 years ago
In Lab 10 you make a stock solution of salicylic acid, and then four dilutions. The stock solution is made by diluting 5.00 ml o
Zigmanuir [339]

Answer:

Stock  solution =  1.25 *10^-3 M

Dilution 1 = 5*10^-4 M

Dilution 2= 3.75 * 10^-4 M

Dilution 3 = 2.5 *10^-4 M

Dilution 4 = 1.25 *10^-4 M

Explanation:

<u>Step 1:</u> Data given

The stock solution is made by diluting 5.00 ml of 1.250 x 10-2 M salicylic acid in 50.00 mL of solution.

<u>Step 2</u>: Calculate the concentration of the stock solution:

M1*V1 = M2*V2

⇒ with M1 = the initial concentration = 1.250 *10^-2 M

⇒ with V1 = 5 mL = 5*10^-3 L

⇒ with M2 = TO BE DETERMINED

⇒ with V2 = 50 mL = 50 *10^-3 L

M2 = (M1*V1)/V2

M2 = (1.250 *10^-2 * 5*10^-3 L) / 50 *10^-3

M2 = 0.00125 M = 1.25 *10^-3 M

<u>Step 3:</u> Calculate dilution 1

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 *10^-3)/(25*10^-3L)

M2 = 0.0005 M = 5*10^-4 M

<u>Step 4</u>: Calculate dilution 2

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 * 7.5*10^-3)/(25*10^-3)

M2 = 0.000375 M = 3.75 * 10^-4 M

<u>Step 5:</u> Calculate dilution 3

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 5*10^-3) /(25*10^-3)

M2 = 0.00025 M = 2.5 *10^-4 M

<u>Step 6</u>: Calculate dilution 4

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 2.5*10^-3)/(25*10^-3)

M2 = 0.000125 M = 1.25 *10^-4 M

5 0
3 years ago
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