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Gemiola [76]
2 years ago
9

What is avogrados number and why is it useful

Chemistry
1 answer:
gtnhenbr [62]2 years ago
7 0

Answer:

Avogadro's number is 6.022×1023 molecules. With Avogadro's number, scientists can discuss and compare very large numbers, which is useful because substances in everyday quantities contain very large numbers of atoms and molecules.

please mark Brainliest if it helps!

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State one reason for the color change in beaker A
myrzilka [38]
Because it either acids or base
4 0
3 years ago
What is the molarity of a 17.0% by mass solution of sodium acetate, NaC2H3O2 (82.0 g/mol), in water? The density of the solution
sattari [20]

Answer:

[NaCH₃COO] = 2.26M

Explanation:

17% by mass is a sort of concentration. Gives the information about grams of solute in 100 g of solution. (In this case, 17 g of NaCH₃COO)

Let's determine the volume of solution, by density

Mass of solution / Volume of solution = Solution density

100 g / Volume of solution = 1.09 g/mL

100 g / 1.09 g/mL = 91.7 mL

17 grams of solute is contained in 91.7 mL

Molarity (M) = Mol of solute /L of solution

91.7 mL / 1000 = 0.0917L

17 g / 82 g/m = 0.207 moles

Molariy = 0.207 moles / 0.0917L → 2.26M

4 0
3 years ago
Which electron configuration represents a Scandium atom?
VARVARA [1.3K]
Sc (neutral) [Ar] 3d1 4s2
Sc+ [Ar] 3d1 4s1
Sc2+ [Ar] 3d1
hope this helped!
4 0
3 years ago
One mole of titanium contains how many atoms
vitfil [10]

6.022 x 10 23 titanium atoms

In 47.88 grams of titanium, there is one mole, or 6.022 x 1023 titanium atoms.

Hope this helps :)

8 0
3 years ago
Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t
Rasek [7]

Answer:

4.6*10^{-14} M

Explanation:

Concentration of Cu^{2+} = [Cu(NO_3)_2] = 0.020 M

Constructing an ICE table;we have:

                                 Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}

Initial (M)             0.020          0.40                        0

Change (M)         -  x                - 4 x                       x

Equilibrium (M)  0.020 -x        0.40 - 4 x              x

Given that: K_f =1.7*10^{13}

K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}

1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}

Since x is so small; 0.40 -4x = 0.40

Then:

1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}

1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}

1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x

8.704*10^9-4.352*10^{11}x =x

8.704*10^9 = 4.352*10^{11}x

x = \frac{8.704*10^9}{4.352*10^{11}}

x = 0.0199999999999540

Cu^{2+}= 0.020 - 0.019999999999954

Cu^{2+} = 4.6*10^{-14} M

8 0
3 years ago
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