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-Dominant- [34]
3 years ago
15

What is the National Suicide rate for Teens?

Chemistry
1 answer:
Butoxors [25]3 years ago
4 0
I think it is 11% I read it on a article on msn.
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There is no connection between the study of chemistry and physics. A. true B. false
9966 [12]
B. False, have a connection between chemistry and physis
4 0
2 years ago
Calculate the standard potential, e∘, for this reaction from its equilibrium constant at 298 k. x(s)+y4+(aq)↽−−⇀x4+(aq)+y(s)k=4.
laila [671]
First, we need to get the number of moles:

from the reaction equation when Y4+ takes 4 electrons and became Y, X loses 4 electrons and became X4+  

∴ the number of moles n = 4

we are going to use this formula:

㏑K = n *F *E/RT

when K is the equilibrium constant = 4.98 x 10^-5

and F is Faraday's constant = 96500

and the constant R = 8.314

and T is the temperature in Kelvin = 298 K

and n is number of moles of electrons = 4 

so, by substitution:

㏑4.98 x 10^-5 = 4*96500*E / 8.314*298

∴E = -0.064 V
6 0
3 years ago
Are moles larger than atoms?
nataly862011 [7]

Answer:

YES BC ATMONS ARE THE SMALLEST UNIT sorry for caps

Explanation:

3 0
2 years ago
Read 2 more answers
Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit
Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

0.0811 M=\frac{n}{0.0017 L}

n = 0.0001378 mol

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0
3 years ago
How would you prepare 3.5 L of a 0.9M solution of KCl?
Fudgin [204]
Calculate the mass of the solute <span>in the solution :

Molar mass KCl = </span><span>74.55 g/mol

m = Molarity * molar mass * volume

m = 0.9 * 74.55 * 3.5

m = 234.8325 g 

</span><span>To prepare 0.9 M KCl solution, weigh 234.8325 g of salt in an analytical balance, dissolve in a beaker, shortly after transfer with the help of a funnel of transfer to a volumetric flask of 100 cm</span>³<span> and complete with water up to the mark, then cover the balloon and finally shake the solution to mix

hope this helps!</span>
8 0
3 years ago
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