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Vera_Pavlovna [14]

2 years ago

13

If a zero order reaction has a rate constant k of 0.0416Mmin and an initial concentration of 2.29 M, what will be its concentrat

ion after 20.0 minutes? Your answer should have three significant figures.

1 answer:

babymother [125]2 years ago

7 0

Answer:

The concentration after 20 mins is 0.832 M

Explanation:

Zero order rate law is given by;

R = K [A₀]⁰

A zero order reaction, rate is independent of the initial concentration

R = K

Where;

R is the rate of reaction

K is the rate constant = 0.0416 M/min

Since R = K,

Then, R = 0.0416 M/min

After 20 min, the concentration will be;

A = Rt

A = (0.0416 M/min)(20 min)

A = 0.832 M

Therefore, the concentration after 20 mins is 0.832 M

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space capsules operate with an oxygen content of about 34%. assuming a total pressure of 780 mm Hg in the space capsule, what is

Lemur [1.5K]

265.2 mmHg is the partial pressure of oxygen in 780 mmHg of total pressure.

Explanation:

The partial pressure of a gas is defined as the individual pressure of the gas in total mixture. In an ideal gas all the constituent gases have partial pressure some of which will give total pressure of the gas.

The partial pressure of a gas is calculated by

total pressure x mole fraction of the gas.

Mole fraction of the oxygen present is 0.34 as it is 34% of the total gas.

$\frac{34}{100}$ = 0.34 is the mole fraction

Total pressure is given as 780 mm Hg

The partial pressure can be calculated using the above formula:

Putting the values in equation:

780 x 0.34

= 265.2 mm Hg is the partial pressure of oxygen.

7 0

2 years ago

A flask of fixed volume contains 1.0 mole of gaseous carbon dioxide and 88 g of solid carbon dioxide. The original pressure and

lianna [129]

Answer:

The correct option is: C. 250 K

Explanation:

Given: <em><u>Before Sublimation-</u></em>

Initial Temperature: T₁ = 300 K, Initial Pressure: P₁ = 1 atm, Initial number of moles of gas: n₁ = 1 mol, given mass of solid Carbon dioxide: w = 88 g

<u><em>After Sublimation- </em></u>

Final Pressure: P₂ = 2.5 atm, Final number of moles of gas: n₂ = ? mol

Final Temperature: T₂ = ? K,

Also, Volume is constant, Molar mass of Carbon dioxide: m = 44 g/mol

As we know,

<em>The number of moles:</em>

$n = \frac {given\: mass\: (w)} {Molar\: mass\: (m)}$

<em>So the number of moles of carbon dioxide sublimed:</em>

$n = \frac {w}{m} = \frac {88\: g} {44\: g/mol} = 2 mol$

<em><u>Therefore, the final number of moles of gas after sublimation:</u></em>

$n_{2} = n_{1} + n = 1\: mol + 2\: mol = 3\: mol$

<u><em>According to the </em></u><u><em>Ideal gas equation</em></u><u><em>:</em></u>

$P.V = n.R.T$

$or, \frac {P_{1}.V_{1}}{n_{1}.T_{1}} = \frac {P_{2}.V_{2}}{n_{2}.T_{2}} \: \: \: \: \: \: ....equation\: (1)$

<em>Since the volume is constant, so the equation (1) can be written as:</em>

$\frac {P_{1}}{n_{1}.T_{1}} = \frac {P_{2}}{n_{2}.T_{2}}$

$\Rightarrow \frac {1\:atm}{1\:mol \times 300\:K} = \frac {2.5\:atm}{3\:mol \times T_{2}}$

$\therefore T_{2} = \frac {2.5\:atm \times 300\:K \times 1\:mol}{3\:mol \times 1\:atm}$

$\Rightarrow T_{2} = 250\:K$

<u>Therefore, the final temperature: T₂ = 250 K</u>

6 0

2 years ago

Brooke decides to model a lunar eclipse. She attaches a large poster of the Sun to her wall to represent the Sun. She then decid

koban [17]

Answer:

its D

Explanation:

7 0

1 year ago

Platinum, which is widely used as a catalyst, has a work function φ(the minimum energy needed to eject an electron from the meta

Arlecino [84]

Answer:

A. $\lambda_0=2.196\times 10^{-7}\ m$

Explanation:

The work function of the Platinum = $9.05\times 10^{-19}\ J$

For maximum wavelength, the light must have energy equal to the work function. So,

$\psi _0=\frac {h\times c}{\lambda_0}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

$\lambda_0$ is the wavelength of the light being bombarded

$\psi _0=Work\ function$

Thus,

$9.05\times 10^{-19}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda_0}$

$\frac{9.05}{10^{19}}=\frac{19.878}{10^{26}\lambda_0}$

$9.05\times \:10^{26}\lambda_0=1.9878\times 10^{20}$

$\lambda_0=2.196\times 10^{-7}\ m$

8 0

2 years ago

How much heat is required to raise the temperature of 670g of water from 25.7"C to 66,0°C? The specific heat

inna [77]

Answer:

Explanation:

q= mc theta

where,

Q = heat gained

m = mass of the substance = 670g

c = heat capacity of water= 4.1 J/g°C

theta =Change in temperature=( 66-25.7)

Now put all the given values in the above formula, we get the amount of heat needed.

q= mctheta

q=670*4.1*(66-25.7)

=670*4.1*40.3

=110704.1

8 0

2 years ago