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3 years ago

PLEASE HELP MEDALS WILL BE AWARDED!!!

1 answer:

4 0

Mole percent of O2 = 10% = 0.1 moles
Mole percent of N2 = 10% = 0.1 moles
Mole percent of He = 80% = 0.8 moles
Molar Mass of O2 = (2 x 16) x 0.1 = 3.2
Molar Mass of N2 = (2 x 14) x 0.1 = 2.8
Molar Mass of He = 4 x 0.8 = 3.2
1. Molar Mass of the mixture = 3.2 + 2.8 + 3.2 = 9.2 grams
2. Since at constant volume density is proportional to mass, so the ratio of
mass will be the ratio of density.
Ratio = Molar Mass of the mixture / Molar Mass of O2 = 9.2 / 32 = 0.2875

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Answer:

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4 years ago

A sample of 0.2140 g of an unkown substance monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.950 M NaOH. Th

morpeh [17]

Solution :

The equation is :

$HA (aq) + NaOH(aq) \rightleftharpoons NaA(aq) + H_2O(l)$

The number of the moles of HA os 0.00285, and the volume is 25 mL.

15 mL of the 0.0950 M NaOH is added.

The total volume of a solution is V = 25 mL + 15 mL = 40 mL

The pH of the solution is 6.50

Calculating the $K_a$ of HA

$HA(aq) \rightleftharpoons A^-(aq)+H^+$

$K_a=\frac{[A^-].[H^+]}{[HA]}$

Let s calculate the concentration of HA and NaOH

$[HA] = \frac{^nH_A}{V}$

$=\frac{0.00285 \ mol}{0.04 \ L}$

= 0.07125 M

$[NaOH]= \frac{0.015L \times 0.0950 M}{V}$

$=\frac{0.001425 mol}{0.04L}$

= 0.0356 M

$HA(aq) \ \ + \ \ NaOH(aq) \ \ \rightleftharpoons NaA(aq) \\ + \ \ H_2O(aq)$

Initial conc. (M) 0.07125 M 0.0356 M 0 M

Change in conc. (M) -0.0356 M -0.0356 M + 0.0356 M

Equilibrium conc. (M) 0.03565 M 0 M 0.0356 M

Therefore, the concentration of HA and the NaA at the equilibrium are [HA] = 0.03565 M and [NaA]= 0.0356 M

0.0356 M of NaA dissociates completely into 0.0356 M $Na^+$ and 0.0356 M $A^-$

Now for $[H^+]$

$[H^+] = 10^{-pH}$

$=10^{-6.5}$

$=3.16 \times 10^{-7}$

Calculating the value of $K_a$ ,

$K_a=\frac{[A^-].[H^+]}{[HA]}$

$=\frac{0.0356 \times 3.16 \times 10^{-7}}{0.03565}$

$=3.16\times 10^{-7}$

Therefore the the value of $K_a$ for the unknown acid is $3.16\times 10^{-7}$ .

4 0

3 years ago

17.0cm

Mademuasel [1]

Answer:

24.68 g/cm³

Explanation:

We'll begin by calculating the volume of the marble. This can be obtained as follow:

Volume = 17 × 23 × 4

Volume = 1564 cm³

Next, we shall convert the mass of the marble (i.e 38.6 kg) to grams (g). This can be obtained as follow:

1 kg = 1000 g

Therefore,

38.6 kg = 38.6 kg × 1000 g / 1 kg

38.6 kg = 38600 g

Therefore, 38.6 kg mass of the marble slab is equivalent to 38600 g.

Finally, we shall determine the density of the marble slab as follow:

Volume = 1564 cm³

Mass = 38600 g.

Density =..?

Density = mass /volume

Density = 38600 g / 1564 cm³

Density = 24.68 g/cm³

Therefore, the density of the marble slab is 24.68 g/cm³

7 0

4 years ago

how do catalysts affect a chemical reaction?(1 point) lower activation energy lower activation energy lose activation energy los

atroni [7]

Catalysts are chemicals that speed up reactions without changing themselves at the conclusion of the process.

By reducing the reaction's activation energy, they accelerate the pace of the reaction. By offering a surface or an orientation, catalysts reduce the activation energy of a process. supplying a different mechanism (a set of procedures the reaction must follow) with lower activation energy. As a result, catalysts facilitate the formation of chemical bonds between atoms to create novel combinations and new compounds.

Learn more about catalysts here:

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6 0

2 years ago

Convert using the relation ship below

olga2289 [7]

Answer:

25.33125KPa

Explanation:

From the above,

1atm = 101.325KPa

0.25atm = (0.25atm x 101.325KPa) / 1atm = 25.33125KPa

Therefore, 0.25atm = 25.33125KPa

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3 years ago