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3 years ago

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How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O)? For example if th

e sample size was 10 L or 10,000 L.

1 answer:

oee [108]3 years ago

5 0

Given :

Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .

To Find :

How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .

Solution :

By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.

So , volume of solution does not matter .

Moles of oxygen , $n=\dfrac{2.666}{16}=0.167\ mole$ .

Now , molecule of CO contains 1 mole of C .

So , moles of C is also 0.167 mole .

Mass of carbon , $m=12\times 0.167=2\ g$ .

Therefore , mass of carbon is 2 grams .

Hence , this is the required solution .

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Now we can use nitrogen as the gas number 2, which travels faster than gas 1;

So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1

We can compare the rates of both the gases;

So here, Rate of gas 2 / Rate of gas 1 = $\sqrt{(molar mass 1 / molar mass 2)}$

Now, 1.687 = square root [ $\sqrt{(molar mass 1) / (28.01 g/mol N_{2})}$ ]

When we square both the sides we get;

2.845 = (molar mass 1) / (28.01 g/mol N2)

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2.845 X (28.01 g/mol N2) = Molar mass 1

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Musya8 [376]

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3 years ago

When 0.250 moles of KCl are added to 200.0 g of water in a constant pressure calorimeter a temperature change is observed. Given

777dan777 [17]

Explanation:

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Therefore, heat generated by dissolving 0.25 moles of KCl will be as follows.

$17.24 kJ/mol \times 0.25 mol$

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or, = 4310 J (as 1 kJ = 1000 J)

Mass of solution will be the sum of mass of water and mass of KCl.

Mass of Solution = mass of water + (no. of moles of KCl × molar mass)

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= 200 g + 13.625 g

= 213.625 g

Relation between heat, mass and change in temperature is as follows.

Q = $mC \Delta T$

where, C = specific heat of water = $4.184 J/g^{o}C$

Therefore, putting the given values into the above formula as follows.

Q = $mC \Delta T$

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3 years ago

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3 years ago

A scientist digs up sample of arctic ice that is 458,000 years old. He takes it to his lab and finds that it contains 1.675 gram

Fiesta28 [93]

Answer:

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Explanation:

Let use the below formula to find the amount of sample

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where

$n = \frac{t}{t_{\frac{1}{2}}}$

here

t = 458,000 years

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$\frac{t}{t_{\frac{1}{2}}}$ = \ $\frac{ 458,000}{229,000}$

n = $\frac{t}{t_{\frac{1}{2}}}$ = 2.000

Now substituting the values

$1.675 = N_0(\frac{1}{2})^{2.000}}$

$1.675 = N_0\times (0.2500)$

$N_0= \frac{1.675}{0.2500}$

$N_0=6.70$

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3 years ago