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ser-zykov [4K]
4 years ago
15

If the lead concentration in water is 1 ppm, then we should be able to recover 1 mg of lead from _____ L of water.

Chemistry
1 answer:
Aliun [14]4 years ago
3 0

Answer:

1 L

Explanation:

ppm means parts per million. Generally the relationship between mass and litre is given as;

1 ppm = 1 mg/L

This means that 1 ppm is equivalent to 1 mg of a substance dissolved in 1 L of water.

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elements are arranged in groups by similar atomic structure on the periodic table. this allows for an elements properties to be
emmasim [6.3K]

Answer:

SUB TO HAVIN RX

Explanation:

8 0
3 years ago
What is the compound MnCl2 called?
givi [52]

Answer is Manganese(ii) chloride

To answer this, you need to know the charges for both Mn and Cl

Mn is in the transition metal category and is unique with having 3 levels of charges

manganese - MN

Manganese(ii) - Mn^2+

Manganese(iii) - Mn^3+


Cl is a nonmetal with the standard charge of Cl^-1

Since you have two charges that don't cancel out, they get switched.

So Cl would receive a positive 2. Because it was a positive 2, we can assume that it starts with Manganese(ii). Then we add chloride to the end of it as it ends with Cl.

8 0
3 years ago
Which energy changes in the enthalpy of solution are endothermic, and which are exothermic?
snow_tiger [21]
When energy is needed in order for the reaction to happen, then that reaction is known as endotermic. When the reaction has as a result energy then it is exotermic. An example of an endotermic reaction would be photosynthesis, for an exotermic: combustion.
8 0
3 years ago
The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC. What is the solubility product for lead(II) iodide?
quester [9]

Answer:

Ksp=1.07x10^{-8}

Explanation:

Hello,

In this case, the dissociation reaction is:

PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)

For which the equilibrium expression is:

Ksp=[Pb^{2+}][I^-]^2

Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)

Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M

In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:

[Pb^{2+}]=1.39x10^{-3}M

[I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M

Thereby, the solubility product results:

Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}

Regards.

6 0
4 years ago
Please help me answer all of these questions! Will give brainliest answer and a heart (probably 5 stars) please and thank you. h
Stels [109]

Answer:

Send me another picture of this so I can help you

5 0
3 years ago
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