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nata0808 [166]
3 years ago
11

Consider the compound Pb(SO4)2 and answer the following questions:

Chemistry
1 answer:
Ivanshal [37]3 years ago
7 0
Hey there ! 

a)  Pb(SO4)2 => <span> lead(II) sulfate 


b) </span><span>The molar mass of this compound was calculated from the amount of atoms of each element, with its respective atomic masses

Pb = 207.2  a.m.u

S = 32.065 a.m.u

O = </span><span>15.9994 a.m.u

Therefore:

</span>Pb(SO4)2 = 207.2 + 32.065 * 2 + 15.9994 * 8 =>  <span>399.3252 g/mol
</span>
hope this helps!
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That is a bar graph.
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2 years ago
Read 2 more answers
(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction
ale4655 [162]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2}   \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu}   \\\\m_{Cu}=0.380gCu

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3}   \\\\m_{AgNO_3}=2.03gAgNO_3

Best regards!

5 0
3 years ago
I need help ASAP!!!!!!!!!!!!!!!!!!!
liraira [26]

last one? don't take my word though

Explanation:

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6 0
1 year ago
How many valence electrons do the halogens possess? group of answer choices
jok3333 [9.3K]

The valence electron does the halogens possess are 7

  • Valence electrons are found in the outermost energy level of an atom
  • They are involved in the formation of chemical bonding with other atoms.
  • The halogens elements are found in group 17 on the periodic table
  • The halogens include fluorine, chlorine, bromine, iodine and astatine.
  • They have seven valence electrons, so they are extremely reactive as they only need one more to fill their outer shell.
  • By octet rule we can say that the electron with 8 outer most shell is full and stable.

Hence the halogens posses 7 valence electron

Learn more about the valence electron on

brainly.com/question/13552988

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5 0
1 year ago
What is the ph of a solution that has a poh of 11.24
galina1969 [7]
Always remember that pH + pOH = 14 
Here, you have a pOH of 11.24, so  you replace it in the equation, and u get:
pH + 11.24 = 14

Then, You move 11.24 to the other part. and moving from a part to another change the sign of the equation. And you get:
pH = 14 - 11.24 = 2.76

So, the pH of a solution that has a pOH of 11.24 is pH = 2.76

Hope this Helps :)
7 0
3 years ago
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