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Alborosie
8 days ago
7

If it takes 15 s for a certain sample of neon to effuse through a porous barrier, how much time will it take for the same amount

of nitrogen gas to effuse through the barrier under the same conditions? 1. greater than 15 s 2. less than 15 s 3. 15 s
Chemistry
1 answer:
alisha [2.9K]8 days ago
5 0

Answer:

The correct response is;

2. takes less than 15 seconds

Explanation:

Graham's law of effusion conveys that

\frac{Rate 1}{Rate2} =\sqrt{\frac{Molar Mass2}{MolarMass 1} }

Thus, knowing that the molar mass of neon gas = 20.18 g/mol

And, the molar mass of nitrogen gas = 28.014 g/mol

We deduce

= 0.72

\frac{Rate 1}{15} =\sqrt{\frac{20.18}{28.014} }So, the effusion rate of nitrogen gas will be 0.72*15 = 12.73 seconds, which is indeed less than 15 seconds.

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Grignard reagents are air-and moisture-sensitive. List at least threereactants, solvents, and/or techniques that were utilized i
Anarel [2989]

Diethyl ether (DTH) and Tetrahydrofuran (THF).

Clarification:

  • Grignard reactions react with water, resulting in the formation of alkanes. The presence of water leads to rapid decomposition of the reagent.
Therefore, solvents like anhydrous diethyl ether or tetrahydrofuran (THF), as well as poly(tetramethylene ether) glycol (PTMG), are used in experimental procedures to limit the exposure of Grignard reagents to air and moisture.
These solvents are chosen because the oxygen they contain stabilizes the magnesium reagent.
THF is a stable compound.
4 0
1 month ago
The phosphorylation of glucose to glucose 6-phosphate Group of answer choices is so strongly exergonic that it does not require
lions [2927]

Answer:

The process of converting glucose to glucose-6-phosphate is an endergonic reaction, which is coupled with the exergonic hydrolysis of ATP.

Explanation:

Within glycolysis, the phosphorylation of glucose to glucose-6-phosphate occurs first, facilitated by the hexokinase enzyme. This reaction is endergonic. This phosphorylation is a coupled reaction tied to ATP hydrolysis, where the free energy released by ATP hydrolysis drives glucose phosphorylation.

5 0
21 day ago
How much CO2 (L) is produced when 2.10 kg of sodium bicarbonate reacts with excess hydrochloric acid at 25.0 °C and 1.23 atm? A)
KiRa [2933]

The equation representing the reaction between sodium bicarbonate and hydrochloric acid is as follows:

NaHCO_3_(_s_) + HCl_(_a_q_) \implies NaCl_(_a_q_) + CO_2_(_g_) + H_2O_(_l_)

The substances NaHCO_3 and HCl combine in a 1:1 ratio. Therefore, we calculate the quantity of sodium bicarbonate and its molar mass to determine the moles formed.

NaHCO_3_M_r = 22.99 + 1.008 + 12.011+ 3 \times 16.0= 84.01 g/mol.

2.1kg\ NaHCO_3 \times \frac{1000g}{kg} \times \frac{mol}{84.01g/mol} = 24.997\ mol.

We also recognize that the stoichiometric proportions are 1:1:1:1:1, which leads to the conclusion that the moles of CO_2 equal 24.977 moles.

Next, we apply the ideal gas equation PV=nRT, where P denotes pressure, V refers to volume, R is the gas constant, and T represents the temperature in kelvins. We rearrange to solve for V

PV= nRT \implies V= \frac{nRT}{P}= \frac{ 24.997\ mol \times 8.2507m^3\ atm \times 298.15K }{mol \times K \times 1.23 atm} = 49967\ m^3

The final answer should be expressed in liters, 1L = 1000\ m^3, hence

49967\ m^3 \times\frac{L}{1000\ m^3} =49.97L\ CO_2\ produced

6 0
1 month ago
Read 2 more answers
The concentration of Si in an Fe-Si alloy is 0.25 wt%. What is the concentration in kilograms of Si per cubic meter of alloy?
KiRa [2933]

Answer: The mass of Si in kilograms is, 19.55kg/m^3

Explanation:

Given that the Si concentration in an Fe-Si alloy is 0.25 weight percent, this translates to:

Mass of Si = 0.25 g = 0.00025 kg

Mass of Fe = 100 - 0.25 = 99.75 g = 0.09975 kg

Density of Si = 2.32g/cm^3=2.32\times 10^6g/m^3

Density of Fe = 7.87g/cm^3=7.87\times 10^6g/m^3

Next, we need to find the quantity of Si in kilograms per cubic meter of alloy.

Si concentration in kilograms = \frac{\text{Weight of Si in 100 g of alloy}}{\text{Volume of 100 g of alloy}}

Si concentration in kilograms = \frac{\text{Weight of Si in 100 g of alloy}}{\frac{\text{Wight of Fe}}{\text{Density of Fe}}+\frac{\text{Wight of Si}}{\text{Density of Si}}}

By substituting all the provided values into this formula, we arrive at:

Si concentration in kilograms = \frac{0.00025kg}{\frac{99.75g}{7.87\times 10^6g/m^3}+\frac{0.25g}{2.23\times 10^6g/m^3}}

Si concentration in kilograms = 19.55kg/m^3

Hence, the mass of Si in kilograms is, 19.55kg/m^3

5 0
1 month ago
Compute 4.659×104−2.14×104. Round the answer appropriately.
lions [2927]

Answer: 25,200.


Explanation:


1) Starting with: 4.659 × 10⁴ - 2.14 × 10⁴


2) Significant figures need to be considered.


Because the powers are equal (10⁴), the decimal values can be subtracted directly. However, it is essential to first check significant figures and the count of decimal places.


3) The figure 4.659 × 10⁴ has four significant digits (4, 6, 5, and 9), whereas 2.14 × 10⁴ contains three significant digits (2, 1, and 4).


4) When adding or subtracting values with a different number of decimal places, the answer must reflect the same number of decimal places as the number with the least amount of decimal precision.


5) Prior to performing subtraction, it is necessary to round the numbers to the least decimal places. Given that 2.14 has two decimal places and 4.659 has three, round 4.659 to 4.66.


6) Now perform the subtraction 4.66 - 2.14 = 2.52


7) Then multiply by the power of 10: 2.52 × 10⁴ = 25,200. This is the final answer.

5 0
24 days ago
Read 2 more answers
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