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Phoenix [80]
3 years ago
7

I need help with the last two questions please help quickly!!

Chemistry
1 answer:
Talja [164]3 years ago
6 0

Not sure about the last one, but I think the percent composition of zinc would be ~ 42.8232

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Calculate the following quantity: molarity of a solution prepared by diluting 45.45 mL of 0.0404 M ammonium sulfate to 550.00 mL
dybincka [34]

Answer:

M_2=3.34x10^{-3}M

Explanation:

Hello!

In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

M_1V_1=M_2V_2

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M

Best regards!

5 0
2 years ago
Do Alkali Metal Compounds form precipitates? And Why?
Alexeev081 [22]

Answer:

Alkali metal hydroxides can be used to test the identity of metals in certain salts. The colour of the precipitate will help identify the metal : Calcium hydroxide is soluble; no precipitate is formed.

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2 years ago
Which pair of elements are the most similar? A. Ca and F B. Na and Cl C. Ne and Ar D. K and Ca
Vinil7 [7]

Answer:

c

Explanation:

8 0
3 years ago
Which phrase correctly describes the outer shell of electrons of each atom in most covalent compounds?
d1i1m1o1n [39]
A, I just took the quiz and that was the answer
8 0
3 years ago
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Please help. thank youuuu
Dmitry [639]

Answer:

'See Explanation

Explanation:

Determine the [OH−] , pH, and pOH of a solution with a [H+] of 9.5×10−13 M at 25 °C.

Given [H⁺] = 9.5 x 10⁻¹³M => [H⁺][OH⁻] = 1.0 x 10⁻¹⁴ => [OH⁻] = 1.0 x 10⁻¹⁴/9.5 x 10⁻¹³ = 0.0105M

pH = -log[H⁺] = -log(9.5 x 10⁻¹³) = - (-1202) = 12.02.

pOH = -log[OH⁻] = -log(0.0105) = -(-1.98) = 1.98

Now you use the same sequence in the remaining problems.

6 0
3 years ago
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