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atroni [7]
3 years ago
6

Which of the following compounds are polar: CH2Cl2, HBr?

Chemistry
1 answer:
Olegator [25]3 years ago
8 0

<em><u>Answer</u></em><em><u> </u></em>: HBr polar

<em><u>please</u></em><em><u> </u></em><em><u>add</u></em><em><u> </u></em><em><u>me</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>brainalist</u></em>

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Given that the ksp of aipo4 is 9.84 x 10-21 at 25°c, calculate the molar solubility of aiop4
svp [43]
AlPO4----> Al+3 + PO4-3

Ksp= [Al+3] x [PO4-3]= 9.84 x 10^-21

Ksp= (x) (x)= x^2
 
X^2= 9.84x10-21
 
x= 9.92 x 10^-11

The molar solubility is 9.92 x 10^-11
8 0
3 years ago
Read 2 more answers
What is economic what is economic ​
irina [24]

Answer:

<h2>ANSWER</h2>
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6 0
3 years ago
What type of atom do you have with p47 and n61
Dmitriy789 [7]
Silver. It is number 47 on the periodic table meaning it has 47 protons. It’s atomic mass is 108. 47+61=108 so it is definitely silver
8 0
3 years ago
What volume in milliliters of 1.420 M sulfuric acid is needed to neutralize 3.209 g of aluminum hydroxide 3 H 2 SO 4 (aq)+2 Al(O
nadya68 [22]

Answer:

V=43.46mL

Explanation:

Hello!

In this case, since the reaction between sulfuric acid and aluminum hydroxide is:

3H_2SO_4+2Al(OH)_3\rightarrow Al_2(SO_4)_3+6H_2O

Whereas the ratio of sulfuric acid to aluminum hydroxide is 3:2; thus, we first compute the moles of sulfuric acid that complete react with 3.209 g of aluminum hydroxide:

n_{H_2SO_4}=3.209gAl(OH)_3*\frac{1molAl(OH)_3}{78.00gAl(OH)_3} *\frac{3molH_2SO_4}{2molAl(OH)_3} \\\\n_{H_2SO_4}=0.0617molH_2SO_4

Then, given the molarity, it is possible to obtain the milliliters as follows:

V=\frac{n}{M}=\frac{0.0617mol}{1.420mol/L}*\frac{1000mL}{1L}\\\\V=43.46mL

Best regards!

7 0
3 years ago
CaC2 + 2H2O → C2H2 + Ca(OH)2 If 2.8 moles of CaC2 are consumed in this reaction, how many grams of H2O are needed?
Strike441 [17]

Answer:

Cac2 is a answer please mark me brainliest

8 0
3 years ago
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