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3 years ago

Which part of the molecule has a partial positive charge? Which part has a partial negative charge?

Chemistry

1 answer:

son4ous [18]3 years ago

3 0

Answer:

partial negative is O(oxygen) and partial positive is (H) Hydrogens

Explanation:

Because the molecule has covalent bonds, all charges are partial.

Partial negative is O(oxygen) and partial positive is (H) Hydrogens.

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If a hard water solution is saturated with calcium carbonate, what volume of the solution has to evaporate to deposit 1.00×102 m

olasank [31]

I think the Ksp for Calcium Carbonate is around 5×10⁻⁹
(I don't know if this is the Ksp value that you use because I read somewhere that this value can vary. You should probably check with your teacher with what Ksp value they want you to use)

the equation for the dissociation CaCO₃ in water is CaCO₃(s)⇄Ca²⁺(aq)+CO₃²⁻(aq) which means that the concentration of Ca²⁺ is equal to the concentration of CO₃²⁻ in solution. For every molecule of CaCO₃ that dissolves, one atom of Ca²⁺ and one molecule of CO₃²⁻ is put into solution which is why the concentrations are equal in solution.

Since Ksp=[Ca²⁺][CO₃²⁻] and we know that [Ca²⁺]=[CO₃²⁻] we can rewrite the equation as Ksp=x² since if you say that [Ca²⁺]=[CO₃²⁻] when you multiply them together you get the concentration squared (I am calling the concentration x for right now).

when solving for x:
5×10⁻⁹=x²
x=0.0000707
Therefore [Ca²⁺]=[CO₃²⁻]=0.0000707mol/L which also shows how much calcium carbonate is dissolved per liter of water since the amount of Ca²⁺ and CO₃²⁻ in solution came from the calcium in a 1 to 1 molar ratio as shown in the equation (the value we found for x is the molar solubility of calcium carbonate).

Using the fact that the molar mass of calcium carbonate is 100.09g/mol you can use dimensional analysis as fallows:
(0.0000707mol/L)(100.09g/mol)=0.007077g/L
That means that there is 0.007077g of Calcium carbonate that can precipitate out of 1L of water.

since the question is asking for how much water needs to be evaporated to precipitate 100mg (0.1g) of Calcium you have to do the fallowing calculation:
(0.1g)/(0.007077g/L)=14.13L of water.
14.13L of water needs to evaporate in order to precipitate out 100mg of calcium carbonate

These types of questions can get long and confusing so I bolded parts that were important to try to guide you through it more easily.

I hope this helps. Let me know if anything is unclear.

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4 years ago

Runoff from agricultural land carries chemicals from fertilizers that collect in a lake. The buildup of chemicals can eventually

bagirrra123 [75]

Answer:

what are the roles of minerals in maintaning homeostasis?

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2 years ago

Which phrase is the best definition for the word common in environmental science?

andreev551 [17]

I believe that the best definition of the word common in environmental science is a typical resource. A typical resource is a resource that is not hard to find and is very common. The answer to the question will hence be A.
The answer could not be the other three choices because a shared resource is a resource that is used by more than one organism in the environment. A worthless resource is one that has no use in the environment while a rare resource is one that is hard to find.

7 0

3 years ago

Read 2 more answers

A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an

erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

7 0

3 years ago

An organic compound that contains a carbonyl group with a hydroxyl group attached to it is an example of a(n)

lawyer [7]

An organic compound that contains a carbonyl group with a hydroxyl group attached to it is an example of a (d) carboxylic acid.

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3 years ago