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ollegr [7]
3 years ago
5

DUE IN 30 MINUTES!!! (please keep it short if you can, if not its ok.) What is the molar mass of Pb(SO4)2? Explain how you calcu

lated this value. (4 points)
Chemistry
2 answers:
timofeeve [1]3 years ago
8 0

Answer:

The molar mass of Pb(SO₄)₂ is 399.32.

Explanation:

To find the molar mass of Pb(SO₄)₂, you have to find the molar mass of each individual element that makes up the compound.

Let's start by figuring out how many of each element is present in the compound.

Pb (lead) - 1

S (sulfur) - 2

O (oxygen) - 8

(SO₄)₂ is a polyatomic ion. It is written in parentheses followed by a subscript of 2. This means that there are two sulfate ions in the compound. So, the number of sulfur atoms and oxygen atoms doubles. (1 × 2 = 2) and (4 × 2 = 8).

Next, look up the molar mass of each element on a periodic table and multiply it by the number of atoms present.

Pb - 207. 2 × 1 = 207. 2

S - 32.06 × 2 = 64.12

O - 16.00 × 8 = 128

Lastly, add up them all together to get the molar mass.

207. 2 + 64.12 + 128 = 399.32

kipiarov [429]3 years ago
5 0

Answer:

399

Explanation:

Pb(SO4)2 contains 1 atom of Pb, 2 atoms of S and 8 atoms of O. So, atomic mass of Pb(SO4)2 is 207 + 64 + 128 = 399 u. Therefore, molar mass of Pb(SO4)2 is 399 g/mol.HOPE THIS helps. Good Luck

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A solution contains 15.27 grams of NaCl in 0.670 kg water at 25 °C. What is the vapor pressure of the solution?
Anvisha [2.4K]

Answer:

The vapor pressure of the solution is 23.636 torr

Explanation:

P_{solution} = X_{solvent}*P_{solvent}

Where;

P_{solution is the vapor pressure of the solution

X_{solvent is the mole fraction of the solvent

P_{solvent is the vapor pressure of the pure solvent

Thus,

15.27 g of NaCl = [(15.27)/(58.5)]moles = 0.261 moles of NaCl

0.67 kg of water =  [(0.67*1000)/(18)]moles = 37.222 moles of H₂O

Mole fraction of solvent (water) = (number of moles of water)/(total number of moles present in solution)

Mole fraction of solvent (water) = (37.222)/(37.222+0.261)

Mole fraction of solvent (water) = 0.993

<u>Note:</u> the vapor pressure of water at 25°C is 0.0313 atm

Therefore, the vapor pressure of the solution = 0.993 * 0.0313 atm

the vapor pressure of the solution = 0.0311 atm = 23.636 torr

5 0
3 years ago
Read 2 more answers
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monitta

Answer:Attemted Failed

Maybe try to search

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5 0
3 years ago
I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0
3 years ago
Mg+o2 balanced eaquation
Tanya [424]

Answer:

2Mg+O2=2MgO

balanced equation

hope it helps.

5 0
3 years ago
Read 2 more answers
Identify the following reaction as oxidation and reduction F2(g)+2e-/2F-(aq)​
sertanlavr [38]

Answer:

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Explanation:

The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

In given reaction fluorine gas gain two electron and form fluoride ions.

F₂(g) + 2e⁻    →   2F⁻(aq)

The given reaction is reduction because oxidation state is decreased from zero to -1.

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