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jonny [76]
3 years ago
14

80cm^3 of oxygen gas diffussed through a porous holebin 50 seconds how long will it take 120cm^3 of nitrogen (iv) oxide to diffu

se through thr same hole under the same confitions​
Chemistry
1 answer:
iogann1982 [59]3 years ago
6 0

Answer:

time for NO₂ to diffuse = 90 s

Explanation:

Data Given:

Amount of O₂ gas transferred = 80 cm³

time of O₂ gas to diffuse = 50s

Amount of NO₂ gas transferred = 120 cm³

time of NO₂ gas to diffuse = ?

Solution:

As we know

  rate of diffusion = Amount of gas transferred / time . . . . . . . (1)

we also know that Graham's Law is

  rate of diffusion gas A/rate of diffusion gas B = \sqrt{mB/mA} . .  . (2)

where

mA = molar mass of gas A

mB = molar mass of gas B

combine both equation 1 and 2

(Amount of gas A transferred / time for gas A) / (Amount of gas B transferred / time for gas B) =  \sqrt{mB/mA} . . . . . . . . (3)

we can write equation 3 for oxygen and nitrogen (iv) oxide

(Amount of gas O₂ transferred / time for gas O₂) / (Amount of gas NO₂ transferred / time for gas NO₂) =  \sqrt{m NO₂/mO₂} . . . . . . . (4)

  • molar mass of O₂ = 2 (16) = 32 g/mol
  • molar mass of NO₂ = 14 + 2(16) = 46 g/mol

Put values in equation equation 4

(80 cm³/50 s) / (120 cm³ / time for NO₂) = \sqrt{46 g/mol/32 g/mol}

(1.6 cm³/s)/(120 cm³ / time for NO₂) = \sqrt{1.44}

(1.6 cm³/s) / (120 cm³ / time for NO₂) = 1.2

Rearrange the above equation

time for NO₂ = 1.2 /(1.6 cm³/s) x 120 cm³

time for NO₂ = 90 s

So,

time for NO₂ to diffuse = 90 s

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It would take 147 hours for 320 g of the sample to decay to 2.5 grams from the information provided.

Radioactivity refers to the decay of a nucleus leading to the spontaneous emission of radiation. The half life of a radioactive nucleus refers to the time required for the nucleus to decay to half of its initial amount.

Looking at the table, we can see that the initial mass of radioactive material present is 186 grams, within 21 hours, the radioactive substance decayed to half of its initial mass (93 g). Hence, the half life is 21 hours.

Using the formula;

k = 0.693/t1/2

k = 0.693/21 hours = 0.033 hr-1

Using;

N=Noe^-kt

N = mass of radioactive sample at time t

No = mass of radioactive sample initially present

k = decay constant

t = time taken

Substituting values;

2.5/320= e^- 0.033 t

0.0078 = e^- 0.033 t

ln (0.0078) = 0.033 t

t = ln (0.0078)/-0.033

t = 147 hours

Learn more: brainly.com/question/6111443

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2 years ago
Suppose you were required to make a solution by diluting 20.0 mL of an aqueous solution to a final volume of 750 mL.
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The first thing you do before performing anything in the laboratory is to read the procedure and prepare the materials needed. Next, if you already have the solution where you are supposed to take your 20 mL sample, then have it near you. Then, prepare a volumetric flask (750 mL) and a 20-mL pipette. Wash the pipette 3 times with the sample solution. If your diluent is water, wash the flask 3 times with water. Now, get 20 mL of sample from your parent solution, then add it to the flask (previously washed with water). Finally, add water until the mark in the flask and make sure that the water added is up to the mark based on the lower meniscus reading to be accurate in the amount inside the flask. <span />
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Hydrogen gas has a density of 0.090 g/L, and at normal pressure and -1.72 C one mole of it takes up 22.4 L. How would you calcul
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Answer:

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Explanation:

Assuming that all caculations are at normal pressure and -1.72°C :

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Where

n is the number of moles of hydrogen

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Chlorine has an atomic number of 17.
zlopas [31]

Answer:

A. 17

Explanation:

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How many moles of calcium carbonate-CaCO3 = 4.15 g​
marin [14]

Answer:

Number of moles = 0.042 mol

Explanation:

Given data:

Number of moles = ?

Mass of calcium carbonate = ?

Solution:

Formula:

Number of moles = mass/ molar mass

now we will calculate the molar mass of calcium carbonate.

atomic mass of Ca = 40 amu

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CaCO₃ = 40 + 12+ 3×16

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Now we will calculate the number of moles.

Number of moles = 4.15 g / 100 g/mol

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