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tresset_1 [31]
3 years ago
12

The average time it takes for a molecule to diffuse a distance of x cm is given by t=x^2/2D where t is the time in seconds and D

is the diffusion coefficient. Given that the diffusion coefficient of glucose is 5.7 x 10^-7 cm2/s, calculate the time it would take for a glucose molecule to diffuse 10 pm, which is roughly the size of a cell.
Chemistry
1 answer:
evablogger [386]3 years ago
5 0

Answer:

Explanation:

time to diffuse x cm is given by the relation

t = X² /2D

D = 5.7 x 10⁻⁷

t = X² / 2 x 5.7 x 10⁻⁷ = X² x 10⁷ / 11.4 s

X = 10 pm = 10 x 10⁻¹² m = 10 x 10⁻¹⁰ cm = 10⁻⁹ cm

t =( 10 ⁻⁹)² x 10⁷ / 11.4 = 10⁻¹¹ / 11.4 = 8.77 x 10⁻¹³ s

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When 8.5 g of methane (ch4) is burned in a bomb calorimeter (heat capacity = 2.677 × 103 j/°c), the temperature rises from 24.00
Scrat [10]
The molar mass of methane is 16 g/mol. The heat absorbed by the calorimeter is the sensible heat, which is the heat gained or lost during a temperature change.

Sensible heat = CΔT = (<span>2.677 kJ/°C)(27.08 - 24</span>°C)
<em>Sensible heat = 8.24 kJ</em>
8 0
3 years ago
Read 2 more answers
Chromium (III) oxide reacts with hydrogen sulfide gas to form chromium (III) sulfide and water. To produce 421 g of cr2s3, how m
maw [93]

Cr₂O₃ ( s ) + 3H₂S ( g ) → Cr₂S₃ ( s ) + 3H₂O ( l )

mol  Cr₂S₃ = 421 : 200.19 g/mol = 2.103

mol Cr₂O₃ ≈ mol Cr₂S₃ = 2.103 ( equivalent coefficient)

mass Cr₂O₃ = 2.103 x 151.99 g/mol = 319.63 gr

6 0
2 years ago
620 mL of nitrogen at standard pressure is compressed into a 480 mL container. What is the new pressure in kPa?
bulgar [2K]

Answer:

c. 131 kPa

Explanation:

Hello!

In this case, since the relationship between volume and pressure is inversely proportional, based on the Boyle's law:

P_1V_1=P_2V_2

Considering that the standard pressure is 101.325 kPa, we can compute the final pressure as shown below:

P_2=\frac{P_1V_1}{V_2}=\frac{620mL*101.325kPa}{480mL}\\\\P_2=131kPa

Therefore, the answer is c. 131 kPa .

Best regards!!

7 0
2 years ago
Which of the following oils would consume the greatest number of equivalents of hydrogen when subject to catalytic hydrogenation
Murljashka [212]

Answer: Option (B) is the correct answer.

Explanation:

When a fatty acid contains high number of double bonds then its unsaturation will also be high and hence, it will consume greater number of equivalents of hydrogen.

In corn oil, there are no unsaturated sites are present.

In olive oil, there is one unsaturated site with majority of oleic acid. In olive oil, there are more than 70% of total unsaturated oils.

In lard oil, there are around 60% of unsaturated oils.

In herring oil, there are highest number of saturated fatty acids and lowest polyunsaturated acids.

Thus, we can conclude that out of the given options, olive oils would consume the greatest number of equivalents of hydrogen when subject to catalytic hydrogenation.

5 0
3 years ago
A gas occupies a volume of 650.0 mL when the pressure is 3.50 atm. What will the new volume be if the pressure is reduced to 1.6
Gemiola [76]

The new volume will be 1379 mL.

Explanation:

As per Boyle's law, the product of initial volume and initial pressure of any gas molecule is equal to the product of final volume and final pressure of those molecules.

So here the initial volume is 650 ml and the initial pressure is 3.50 atm. As the temperature is said to be constant, then this system will be obeying Boyle's law. So, the final pressure is given as 1.65 atm. As there is a reduction in the pressure, the volume of the gas is tend to get expanded.

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

So, (650*10^{-3}*3.50)=(1.65*V_{2})

V_{2} = \frac{650*10^{-3}*3.50}{1.65} = 1379 mL

So, the new volume of the gas on reduction in pressure is 1379 mL.

7 0
3 years ago
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