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2 years ago

2. (05.02 MC)

Chemistry

1 answer:

iVinArrow [24]2 years ago

5 0

Solution :

We know, molecular mass of NH₃ is , M = 17 g/mol .

Number of moles of NH₃ in 7.2 × 10²⁴ are :

$n = \dfrac{7.2 \times 10^{24}}{6.022 \times 10^{23}}\\\\n = 11.96\ moles$

So, mass of NH₃ present in 11.96 moles is :

m = 11.96 × 17 gram

m = 203.32 gram

Hence, this is the required solution.

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What is the final volume (L) of a 1.00 L system at 315 K and 1.10 atm if STP conditions are established?

LenaWriter [7]

Answer:

0.95L

Explanation:

Data obtained from the question include:

V1 (initial volume) = 1L

T1 (initial temperature) = 315K

P1 (initial pressure) = 1.10 atm

T2 (final temperature) = stp = 273K

P2 (final pressure) = stp = 1atm

V2 (final volume) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final volume of the system can be obtained as follow:

P1V1/T1 = P2V2/T2

1.1 x 1/315 = 1 x V2/273

Cross multiply to express in linear form.

315 x V2 = 1.1 x 273

Divide both side by 315

V2 = (1.1 x 273) /315

V2 = 0.95L

Therefore, the final volume of the system if STP conditions are established is 0.95L

5 0

2 years ago

The concentration of a NaOH solution was experimentally determined by dissolving 0.7816 grams KHP (204.2212 grams/mol) in 50.0 m

REY [17]

Answer:

The concentration of the standard NaOH solution is 0.094 moles/L.

Explanation:

In the titration, the equivalence point is defined as the point where the moles of NaOH (the titrant) and KHP (the analyte) are equal:

moles of NaOH = moles of KHP

$[NaOH]xV_{NaOH} = moles of KHP$

$[NaOH] = \frac{moles of KHP}{V_{NaOH}}$

The $V_{NaOH}$ is 40.82mL = 0.04082L and the moles of KHP are

$0.7816g / 204.2212\frac{g}{mol} = 3.827x10^{-3} moles$

Replacing at the first equation:

$[NaOH] = \frac{3.827x10^{-3}moles}{0.04082L} = 0.094 moles/L$

3 0

2 years ago

Consider the following reaction.

Deffense [45]

The pressure of NO2 and N2O4 would be 1.756 atm and 0.1542 atm respectively.

According to Boyle's law, the pressure of a gas is inversely proportional to its volume at a constant temperature.

Mathematically: P1V1=P2V2

When the volume of a gas is halved at a constant temperature, the pressure is doubled. If it is the volume is doubled, the pressure would be halved.

In this case, the volume of the container is halved. Meaning that the pressure of each of the gas would be doubled by the time the equilibrium is reestablished.

Thus, NO2 would double from 0.878 atm to 1.756 atm while N2O4 will double from 0.0771 atm to 0.1542 atm

More on gas laws can be found here: brainly.com/question/1190311

8 0

2 years ago

Calculate the pH of a buffer solution made by adding 15.0 g anhydrous sodium acetate (NaC2H3O2) to 100.0 mL of 0.200 M acetic ac

DiKsa [7]

Answer:

pH of Buffer Solution 5.69

Explanation:

Mole of anhydrous sodium acetate = $\frac{Given mass}{Molecular mass}$

= $\frac{15}{82}$

= 0.18 mole

100 ml of 0.2 molar acetic acid means

= M x V

= 0.2 x 100

= 20 mmol

= 0.02 mole

Using Henderson equation to find pH of Buffer solution

pH = pKa + log $\frac{[Salt]}{[Acid]}$

= 4.74 + log $\frac{0.18}{0.02}$

= 4.74 + log 9

= 5.69

So pH of the Buffer solution = 5.69

8 0

2 years ago

0.0524 moles of H2O equals how many molecules of water?

Charra [1.4K]

Hey there!:

<span>Use the Avogadro constant :
</span>
1 mole -------------- 6.02*10²³ molecules
0.0524 moles -------- molecules H2O

molecules H2O = 0.0524* ( 6.02*10²³) / 1

molecules H2O = 3.15*10²² / 1

= 3.15*10²² molecules H2O

8 0

2 years ago

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