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3 years ago

3 Classify the following into elements , compounds and mixtures.

Chemistry

1 answer:

Natalija [7]3 years ago

6 0

gold:element

salt solution: mixture

bronze: compound

calcium carbonate: compound

hydrogen: element

soil: mixture

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It took 4.5 minutes for 1.0 l helium to effuse through a porous barrier. how long will it take for 1.0 l cl2 gas to effuse under

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When The rate of effusion is inversely proportional to the √molar mass of the substance.
and we have R(He) = 1L / 4.5 min so,
R(He)/R(Cl2) = (molar mass of Cl2/ molar mass of He)^0.5
and when we have the molar mass of Cl2 = 70.9 & the molar mass of He = 4
so by substitution:
(1L/4.5 min)/ R(Cl2) = (70.9 / 4)^0.5
(1L/4.5 min) / R(Cl2) = 4.21
∴R(Cl2) = (1L/4.5 min) / 4.21 = 1L/ (4.5*4.21)min = 1 L / 18.945 min
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4 years ago

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Carbon monoxide (CO) is a poisonous gas because it binds very strongly to the oxygen carrier hemoglobin in blood. A concentratio

velikii [3]

<u>Answer:</u> The amount of CO that is occupied in the room is $1.98\times 10^3L$

<u>Explanation:</u>

We are given:

Concentration of CO = $8.00\times 10^2ppm=800pm$ by volume

This means that $800\mu L\text{ or }800\times 10^{-6}$ of CO is present in 1 L of blood

To calculate the volume of cuboid, we use the equation:

$V=lbh$

where,

V = volume of cuboid

l = length of cuboid = 10.99 m

b = breadth of cuboid = 18.97 m

h = height of cuboid = 11.89 m

$V=10.99\times 18.97\times 11.89=2478.83m^3$

Converting this into liters, by using conversion factor:

$1m^3=1000L$

So, $2478.83m^3=2.479\times 10^6L$

Applying unitary method:

In 1 L of blood, the amount of CO present is $800\times 10^{-6}$

So, in $2.479\times 10^6L$ of blood, the amount of CO present will be = $\frac{800\times 10^{-6}}{1}\times 2.479\times 10^{6}=1983.2L=1.98\times 10^3L$

Hence, the amount of CO that is occupied in the room is $1.98\times 10^3L$

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