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4 years ago

What is potential energy?

2 answers:

4 0

The energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors.

Helga [31]4 years ago

4 0

Answer:

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Explanation:

Potential energy is the energy stored in a body due to the body’s positioning or state.

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The chemistry of anhydrides is similar to that of acid chlorides, although they react more slowly.

Volgvan

Answer:

both

Explanation:

6 0

3 years ago

A 5.00g quantity of a diprotic acid was dissolved in water and made up exactly 250 mL. Calculate the molar mass if the acid is 2

MAXImum [283]

Answer:

The molar mass of the diprotic acid is 90.10 g/mol.

Explanation:

The acid is 25.0 mL of this solution required 11.1 mL of 1.00 KOH for neutralization.

$H_2A+2KOH\rightarrow K_2A+2H_2O$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$ ( neutralization )

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of diprotic acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=?\\V_1=25 mL\\n_2=1\\M_2=1.00 M\\V_2=11.1 mL$

Putting values in above equation, we get:

$2\times M_1\times 25=1\times 1.00\times 11.1}$

$M_1=\frac{1\times 1.00\times 11.1}{2\times 25}=0.222 M$

Molarity of acid solution = 0.222 M =0.222 mol/L

Volume of original solution = 250 mL = 0.250 L ( 1 mL = 0.001 L)

Moles of diprotic acid in 0.250 L solution :

$=0.222 mol/L\times 0.250 L=0.0555 mol$

Mass of diprotic acid = m = 5.00 g

$Moles(n)=\frac{mass(m)}{\text{Molar mass(M)}}$

$M=\frac{m}{n}=\frac{5.00 g}{0.0555 mol}=90.10 g/mol$

The molar mass of the diprotic acid is 90.10 g/mol.

5 0

3 years ago

A crystalline solid has a high melting point and is known to be held together with covalent bonds. This solid is an example of _

OverLord2011 [107]

Letter B is correct the answer

5 0

4 years ago

In standardizing the solution of aqueous sodium hydroxide, a chemist overshoots the end point and adds too much naoh(aq). would

irinina [24]

Answer: Concentration of NaOH calculated will be underestimated.

Explanation:

End point is an observational point , which tells us about the completion of reaction between the titrant (solution in burette) and titre(solution in conical flask) in titration experiment.

In this case , NaOH is titrant whose concentration is unknown.

$M_1=\text{molarity of titre}$ , $M_2=\text{molarity of NaOH}$

$V_1=\text{volume of titre}$ , $V_2=\text{volume of NaOH}$

$M_1V_1=M_2V_2$

$M_2=\frac{M_1V_1}{V_2}$ ....(1)

According to question a chemist overshoots the end point and adds to much of NaOH solution, which means increase in the value of $V_2$ .

Then the value of $M_2$ in equation (1), will get lowered , which means that the concentration of NaOH was lower than that of the actual value. Hence underestimated concentration of NaOH.