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3 years ago

Why are covalent bonds between hydrogen and nitrogen or oxygen polar? see section 2.1 ( page 57) ?

1 answer:

5 0

A covalent bond is on in which electron pairs are shared between one or more electron pairs between atoms. Polar covalent bonds are so called because the electron density is closer to one atom participating in the bond than to another. The tendency of electron density to be closer to one atom and not the other can be due to the geometry of the bond.

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When water reaches its boiling point and turns into water vapor what happens to its molecular structure

Alex787 [66]

When water reaches its boiling point and turns into water vapor, the molecular structure of water remains the same. It is only the state of the substance that is changed in this process. I hope this helps you

8 0

3 years ago

!!!!!PLEASE HELP!!!!!

ollegr [7]

Answer:

Difference in the potential energy of the reactants and products

Explanation:

The products have a lower potential energy than the reactants, and the sign of ΔH is negative. In an endothermic reaction, energy is absorbed. The products have a higher potential energy than the reactants, and the sign of ΔH is positive.

3 0

2 years ago

Who else had to do a final model for the rust unit

ArbitrLikvidat [17]

Which one? Are u saying.

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2 years ago

in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc

Margarita [4]

Answer: 0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$ .....(1)

a) Molarity of $Na_2S$ solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

$0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol$

$\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol$

b) Molarity of $Hg(NO_3)_2$ solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

$0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol$

$Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3$

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ reacts with 1 mole of $Na_2S$

Thus 0.0005 moles of $HgNO_3$ reacts with= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg(NO_3)_2$

Thus $Hg(NO_3)_2$ is the limiting reagent and $Na_2S$ is the excess reagent.

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ forms= 1 mole of $Hg_2S$

Thus 0.0005 moles of $Hg(NO_3)_2$ forms= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg_2S$

mass of $H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g$

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0

3 years ago

What is the molarity of a solution in which 175.8<br> grams of NaCl is dissolved in 1.5 L of water?

Sophie [7]

175.8 g NaCl to moles:
(175.8 g)/(58.44 g/mol) = 3.008 mol NaCl

Molarity = moles of solute/volume of solution in liters

(3.008 mol NaCl)/(1.5 L) = 2.0 M.

The molarity of this solution would be 2.0 M.

8 0

2 years ago