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3 years ago

What happens to the particles during solid to a liquid

Chemistry

1 answer:

Ratling [72]3 years ago

3 0

the particles from solid to liquid start to move around faster then it was at the first state

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Chemical equation for cycloheptane (l)

Roman55 [17]

Answer: C7H14 - PubChem.

Explanation:

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3 years ago

QUICK I JUST NEED THE ANSWER!!! What's the hybridization of the central atom in OF2?

Blababa [14]

Answer:

sp3 hybridization

Explanation:

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3 years ago

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3 years ago

A researcher was attempting to quantify the amount of dichlorodiphenyltrichloroethane (DDT) in spinach with gas chromatography u

devlian [24]

Answer:

0.0136mg DDT / g spinach

Explanation:

Quantification in chromatography by internal standard has as formula:

RF = Aanalyte×Cstd / Astd×Canalyte (1)

Where RF is response factor, A is area and C is concentration

Replacing with first experiment values:

RF = 5019×3.20mg/L / 8179×6.37mg/L

RF = 0.308

In the next experiment, final concentration of chloroform was:

11.45mg/L × (1.25mL / 25.00mL) = 0.5725mg/L

From (1), it is possible to write:

Aanalyte×Cstd / Astd×RF = Canalyte

Replacing:

6821×0.5725mg/L / 14061×0.308 = Canalyte

Canalyte = 0.9017mg/L

as the sample was made from 0.750mL of extract. Concentration of extract is:

0.9017mg/L × (25.00mL / 0.750mL) = 30.06mg/L. As the extract has a volume of 2.40mL:

30.06mg/L × 2.40x10⁻³L = 0.07213mg of DDT in the extract

As the extract was made from 5.29g of spinach:

0.07213mg of DDT in the extract / 5.29g spinach = 0.0136mg DDT / g spinach

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4 years ago

Calculate the ph at the equivalence point for the titration of 0.230 m methylamine (ch3nh2) with 0.230 m hcl. the kb of methylam

Vlad1618 [11]

Answer: The pH at the equivalence point for the titration will be 0.65.

Solution:

Let the concentration of $[OH^-]$ be x

Initial concentration of $[CH3NH_2]$ , c = 0.230 M

$CH_3NH_2+H_2O\rightleftharpoons CH_3NH_3^++OH^-$

at eq'm c-x x x

Expression of $K_b$ :

$K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}$

Since ,methyl-amine is a weak base,c>>x so $c-x\approx c$ .

$K_b=\frac{x^2}{c}=5.0\times 10^{-4}=\frac{x^2}{0.230 M}$

Solving for x, we get:

$x=1.07\times 10^{-2} M$

Given, HCl with 0.230 M , it dissociates fully in water which means $[H^+]$ = 0.230 M

$[OH^-]=[H^+]$ will result in neutral solution, since $[OH^-]$

Remaining $[H^+]$ after neutralizing $[OH^-]$ ions

$[H^+]_{\text{left in solution}}=[H^+]-[OH^-]=0.230-1.07\times 10^{-2}=0.2193 M$

$pH=-log{[H^+]_{\text{left in solution}}=-log(0.2193)=0.65$

The pH at the equivalence point for the titration will be 0.65.

8 0

4 years ago