All categories

2 years ago

If there is initially 1.334 m o2, how long will it take for the o2 concentration to reach 0.443 m?

1 answer:

3 0

To answer, we take it that the rate of the consumption of O2 in this item is directly proportional to its concentration in first degree such that,
-dCa/dt = kCa
Solving,
-dCa/Ca = kt
lnCa - lnCao = -kt
Substituting,
ln(0.443) - ln(1.334) = -k(t)
t = 1.1023/k

You might be interested in

How would you obtain ethanol from rice starch?

sergejj [24]

Answer:

By putting the rice in water

Explanation:

5 0

2 years ago

A container has a volume of 2.79 L and a pressure of 5.97 atm. If the pressure changes to 1460 mm Hg, what is the container’s ne

maw [93]

Answer:

8.68 L is the new volume

Explanation:

You use Boyle's law for this.

$P_{1}V_{1}=P_{2}V_{2}$

$P_{1}$ = first pressure

$P_{2}$ = second pressure

$V_{1}$ = first volume

$V_{2}$ = second volume

Convert pressure from atm to mmHg (use same units):

5.97 x 760 = 4537.2 -> 4.54 x 10³

...maintain 3 significant figures in calculation, and round as needed...

(4.54 x 10³ mmHg)(2.79 L) = (1460 mmHg)( $V_{2}$ )

(4.54 x 10³ mmHg)(2.79 L) / (1460 mmHg) = $V_{2}$ = 8.68 L

Hope this helps :)

5 0

2 years ago

How many ml of concentrated hydrochloric acid would be required to make 1 L of a 0.2 M solution? Assume that concentrated hydroc

fgiga [73]

<u>Answer:</u> The volume of concentrated hydrochloric acid required is 16.53 mL

<u>Explanation:</u>

To calculate the volume of concentrated solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution

We are given:

Conversion factor: 1 L = 1000 mL

$M_1=12.1M\\V_1=?mL\\M_2=0.2M\\V_2=1L=1000mL$

Putting values in above equation, we get:

$12.1\times V_1=0.2\times 1000\\\\V_1=16.53mL$

Hence, the volume of concentrated hydrochloric acid required is 16.53 mL

3 0

2 years ago

The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C

artcher [175]

Explanation:

The given data is as follows.

$P_{1}$ = 100 mm Hg or $\frac{100}{760}atm$ = 0.13157 atm

$T_{1}$ = $1080 ^{o}C$ = (1080 + 273) K = 1357 K

$T_{2}$ = $1220 ^{o}C$ = (1220 + 273) K = 1493 K

$P_{2}$ = 600 mm Hg or $\frac{600}{760}atm$ = 0.7895 atm

R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

$log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}$

$log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]$

$log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]$

0.77815 = $\frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K$

$\Delta H_{vap}$ = $2.219 \times 10^{5}$ J/mol

= $2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}$

= 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0

2 years ago

whg doe sthe transfer of electrons from magnesium to oxygen cause the ions produced to attract eachother

almond37 [142]

Oxygen gains two electrons when it bonds to form a complete outer shell and magnesium loses two electrons when bonding to gain its full outer shell.

As electrons are negative, the oxygen (which gains electrons) will become negative and the magnesium (which loses electrons) will become positive.

The negative and positive ions will then attract to one another due to the magnetic pull of the positive and negative.

8 0

2 years ago