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3 years ago

8

An attempt to explain observations of the natural world is

1 answer:

atroni [7]3 years ago

3 0

A pseudoscientific theory

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Calculate the [OH-] and the pH of 0.035 M KOH.

DanielleElmas [232]

Answer:

pH

$= 12.54$

OH- concentration 28.84

Explanation:

KOH dissociates into K+ and OH-. The ratio of K+ and OH- ion is 1:1

In any aqueous solution, the H3O+ and OH - must satisfy the following condition -

$[ H_3O^+] [OH^-] = k_w$

$[ H_3O^+] = \frac{k_w}{ [OH^-]}$

$[H_3O^+] = \frac{1 * 10^{-14}}{3.5 *10^{-2}[H_3O^+] = 2.857 * 10^{-13}$ M

pH =

$- log [ H_3O^+]\\- log [2.857 * 10^{-13}]$

pH

$= - [-12.54]\\= 12.54$

pOH $= 14 -$ pH

pOH

$= 14 - 12.54\\= 1.46$

OH concentration $= antilog 1.46$

OH- concentration 28.84

5 0

4 years ago

Two unknown household items are being tested with litmus paper. Substance A turns red litmus paper blue and Substance B turns bl

irina1246 [14]

Answer:

Substance A is a base and Substance B is an acid.

According to the experimental results

7 0

3 years ago

How much energy would be absorbed as heat by 75 g of iron when heated from 295 k to 301 k?

Katen [24]

Answer:

$Q=202.5J$

Explanation:

Hello,

In this case, the increase in the temperature involves the addition of heat that is defined in terms of mass, heat capacity and temperature:

$Q=mCp(T_2-T_1)$

In this case, the heat capacity of iron is 0.450 J/(g*K), thus the heat results:

$Q=75g*0.450\frac{J}{g*K}*(301-295)K\\\\Q=202.5J$

In such a way, since the temperature is increased heat is added, that is why it is positive.

Best regards.

4 0

4 years ago

Read 2 more answers

In the periodic table, a set of properties repeats from

jok3333 [9.3K]

in the periodic table, a set of properties repeats from row to row

3 0

3 years ago

Read 2 more answers

Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0

klemol [59]

Answer:

Partial pressure of methane: 1.18 atm

Partial pressure of ethane: 1.45 atm

Partial pressure of propane: 2.35 atm

Explanation:

Let the total moles of gases in a container be n.

Total pressure of the gases in a container =P = 5.0 atm

Temperature of the gases in a container =T = 23°C = 296.15 K

Volume of the container = V = 10.0 L

$PV=nRT$ (Ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.0 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.0564 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16.04 g/mol}=0.4878 mol$

Moles of ethane gas = $n_2=\frac{18.00 g}{30.07 g/mol}=0.5986 mol$

Moles of propane gas = $n_3=?$

$n=n_1+n_2+n_3$

$n_3=n-n_1-n_2=2.0564 mol-0.4878 mol-0.5986 mol= 0.9700 mol$

Partial pressure of all the gases can be calculated by using Raoult's law:

$p_i=P\times \chi_i$

$p_i$ = partial pressure of 'i' component.

$\chi_1$ = mole fraction of 'i' component in mixture

P = total pressure of the mixture

Partial pressure of methane:

$p_1=P\times \chi_1=P\times \frac{n_1}{n_1+n+2+n_3}=P\times \frac{n_1}{n}$

$p_1=5.00 atm\times \frac{0.4878 mol}{2.0564 mol}=1.18 atm$

Partial pressure of ethane:

$p_2=P\times \chi_2=P\times \frac{n_2}{n_1+n+2+n_3}=P\times \frac{n_2}{n}$

$p_2=5.00 atm\times \frac{0.5986 mol}{2.0564 mol}=1.45 atm$

Partial pressure of propane:

$p_3=P\times \chi_3=P\times \frac{n_3}{n_1+n+2+n_3}=P\times \frac{n_3}{n}$

$p_3=5.00 atm\times \frac{0.9700 mol}{2.0564 mol}=2.35 atm$

5 0

3 years ago