Semasa proses pengewapan, cecair bertukar menjadi gas. Apakah yang berlaku kepada zarah dalam cecair??

What is the boiling point of a 0.75-molal solution of a non-volatile non-electrolyte solute in water? Kb for water = 0.52 degree

AVprozaik [17]

Take the molality and multiply it by the Kb
0.75 × 0.52 = 0.39 C

then add the boiling pt to the 0.39
100 + 0.39 = 100.39 C

3 0

3 years ago

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How much pv work is done in kilojoules for the reaction of 0.68 mol of h2 with 0.34 mol of o2 at atmospheric pressure if the vol

dsp73

The delta H of -484 kJ is the heat given off when 2 moles of H2 react with 1 mole of O2 to make 2 moles of H2O. You don't have anywhere near that much reactants, only 1/4 as much

<span>actual delta H = 0.34 moles H2 x (-484 kJ / 2 moles H2) = 823 kJ </span>

<span>delta E = delta H - PdeltaV = 823 kJ - 0.41 kJ = 822 kJ</span>

3 0

3 years ago

Explain whether nitrogen atoms will form bonds with other atoms

iren2701 [21]

The closer the atoms are the more the nitrogen will be come bigger or the oppisite

3 0

2 years ago

The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentrat

Valentin [98]

Answer:

(a)

$0.0342M$

(b)

$t_{1/2}=17.36s\\t_{1/2}=23.15s$

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

$\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\$

$[NOBr]=\frac{1}{29.2/M}=0.0342M$

(b) Now, for a second-order reaction, the half-life is computed as shown below:

$t_{1/2}=\frac{1}{k[NOBr]_0}$

Therefore, for the given initial concentrations one obtains:

$t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.072M}=17.36s\\t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.054M}=23.15s$

Best regards.

8 0

2 years ago

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3) Lithium metal (Li) react with hydrosulfuric acid (HS) to produce hydrogen gas and magnesium chloride (Li 2 S) . How many gram

Vaselesa [24]

Answer:

3.066g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2Li + H2S —> Li2S + H2

Step 2:

Data obtained from the question.

Volume (V) of H2 = 4.5L

Temperature (T) = 315K

Pressure (P) = 1.258 atm

Note:

Gas constant (R) = 0.0821atm.L/Kmol

Number of mole (n) of H2 =..?

Step 3:

Determination of the number of mole of H2 produced.

This can be obtained by using the ideal gas equation as follow

PV = nRT

Divide both side by RT

n = PV /RT

n = 1.258 x 4.5 / 0.0821 x 315

n = 0.219 mole

Therefore, 0.219 mole of H2 is produced.

Step 4:

Determination of the number of mole of Li that will produce 0.219 mole of H2.

This is shown below:

2Li + H2S —> Li2S + H2

From the balanced equation above,

2 moles of Li reacted to produce 1 mole of H2.

Therefore, Xmol of Li will react to produce 0.219 mole of H2 i.e

Xmol of Li = 2 x 0.219

Xmol of Li = 0.438 mole

Step 5:

Conversion of 0.438 mole of Li to grams.

Number of mole of Li = 0.438 mole

Molar Mass of Li = 7g/mol

Mass = number of mole x molar Mass

Mass of Li = 0.438 x 7

Mass of Li = 3.066g

Therefore, 3.066g if Li is needed for the reaction.

6 0

3 years ago