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Scilla [17]
3 years ago
12

What is the name of the compoud H2Co2

Chemistry
1 answer:
Irina18 [472]3 years ago
7 0
The answer is carbonous acid periodt periodt periodt periodt period

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PLEASE PLEASE PLEASE​ ANSWER HAVE MERCY 35 POINTS
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You should give answers since some people will just try and find any answer, which happened to me. So like if you have answers for the question just put them like A B C D kinda like true or false for others to know what they got instead of giving them more work, and then they just give you random answers.
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3 years ago
Conclusion: Summarize the results of this experiment in one or two sentences. What
GaryK [48]

Answer:

The flame test is used to visually determine the identity of an unknown metal or metalloid ion based on the characteristic color the salt turns the flame of a Bunsen burner.

4 0
3 years ago
Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /
barxatty [35]

Answer:

[Pb^{2+}]=3.9 \times 10^{-2}M

this is the concentration required to initiate precipitation

Explanation:

PbCl_2  ⇄ Pb^{2+}+2Cl^-

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

Ip=[Pb^{2}][2Cl^-]^2=Ksp

Ksp=2.4\times 10^{-4}

lets solubility=S

[Pb^{2+}] = S

[Cl^-]=2S

Ksp=[Pb^{2+}]\times [Cl^-]^2

Ksp=S \times (2S)^2

Ksp=4S^3

S=\sqrt[3]{\frac{Ksp}{4} }

S=3.9\times 10^{-2}

[Pb^{2+}]=3.9 \times 10^{-2}M this is the concentration required to initiate precipitation

4 0
3 years ago
When solutions of silver nitrate and sodium chloride are mixed, silver chloride precipitates out of the solution according to th
kipiarov [429]

The concentration of the sodium chloride would be 0.082 M

<h3>Stoichiometric calculations</h3>

From the equation of the reaction, the ratio of AgCl produced to NaCl required is 1:1.

Mole of 46.6 g AgCl produced = 46.6/143.32 = 0.325 moles

Equivalent mole of NaCl = 0.325 moles.

Molarity of 0.325 moles, 3.95 L NaCl = mole/volume = 0.325/3.95 = 0.082 M

More on stoichiometric calculations can be found here: brainly.com/question/27287858

#SPJ1

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2 years ago
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