Ask question

Ask question

All categories

2 years ago

8

400 mL of hydrogen are collected over water at 18 C and a total pressure of 740 mm of mercury. (Vapor pressure of H2O at 18 C=15

.5mm).
a) What is the partial pressure of H2O?

b) What is the partial pressure of H2?

1 answer:

MrRa [10]2 years ago

6 0

Answer:

a)15.5 mmHg is the partial pressure of water vapor.

b) 724.5 mmHg is the partial pressure of hydrogen gas.

Explanation:

Total pressure of gases = T = P = 740 mmHg

Vapor pressure of the water = $p_1=15.5 mmHg$

Partial pressure of the hydrogen gas = $p_2=?$

Partial pressure of the water = $p_1=15.5 mmHg$

15.5 mmHg is the partial pressure of water vapor.

Using Dalton's law of partial pressure:

$P=p_1+p_2$

$740 mmHg=15.5 mmHg+p_2$

$p_2=740 mmHg-15.5mmHg =724.5 mmHg$

724.5 mmHg is the partial pressure of hydrogen gas.

You might be interested in

What molecule is the ultimate source of hydrogen ions that are secreted into the gastric juice?

sladkih [1.3K]

Answer: carbonic acid

Explanation:

8 0

2 years ago

Which of these species are free radicals? check all that apply. check all that apply. no2 no o o2 o3?

professor190 [17]

Answer: NO2, NO, and O2.

<span>Free radicals are toxic substances produced by the body. In normal circumstances,the body can neutralize but<span> when the level of these substances is to much,they accumulate and can generate diseases, such as osteoporosis and cancer.</span></span>

4 0

2 years ago

Like all equilibrium constants, Kw varies somewhat with temperature. Given that Kw is 3.31 × 10−13 at some temperature, compute

Artyom0805 [142]

Since Kw= [H⁺][OH⁻], and the concentration of both substances are the same, the equation is now Kw=[H⁺]²
So,
3.31x10⁻¹³ = [H⁺]²
Take the square root= 5.75x10⁻⁷
Then take the negative log to find the pH:
-log(5.75x10⁻⁷) = 6.25

5 0

2 years ago

Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b

frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ ?

(a) $Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}$

(b) $2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}$

(c) $Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}$

(d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Answer: (d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Explanation: <u>Redox</u> <u>Reaction</u> is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called <u>cell</u> <u>notation,</u> formed by: <em><u>left side</u></em> of the salt bridge (||), which is always the <em><u>anode</u></em>, i.e., its half-equation is as an <em><u>oxidation</u></em> and <em><u>right side</u></em>, which is always <em><u>the cathode</u></em>, i.e., its half-equation is always a <em><u>reduction</u></em>.

For the cell notation: $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$

Aluminum's half-equation is oxidation:

$Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}$

For Lead, half-equation is reduction:

$Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}$

Multiply first half-equation for 2 and second half-equation by 3:

$2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}$

$3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}$

Adding them:

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

The balanced redox reaction with cell notation $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ is

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

6 0

2 years ago

A student uses visible spectrophotometry to determine the concentration of CoCl2(aq) in a sample solution. First the student pre

k0ka [10]

Answer:

$4\times 10^{-9} J$ is the approximate energy of one photon of this light.

Explanation:

Energy of the photon can be calculated by

$E=h\nu=\frac{h\times c}{\lambda}$ (Planck's equation)

where,

E = energy of photon

h = Planck's constant = $6.63\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of light =

$\nu$ = frequency of the light

we have , $\lambda =510 nm=510\times 10^{-9}m$

Now put all the given values in the above formula, we get the energy of the photons.

$E=\frac{(6.63\times 10^{-34}Js)\times (3\times 10^8m/s)}{510\times 10^{-9}m}$

$E=3.9\times 10^{-19}J\approx 4\times 10^{-9} J$

$4\times 10^{-9} J$ is the approximate energy of one photon of this light.

5 0

2 years ago