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3 years ago

How many grmas of CaCO3 are present in a sample if there are 4. 52 x 10^24 atoms of carbon in that sample?

Chemistry

1 answer:

lorasvet [3.4K]3 years ago

3 0

No of atoms in 1 mol of CaCO3=6.023×10^23

$\\ \sf\bull\longmapsto No\:of\:moles=\dfrac{4.52\times 10^24}{6.023\times 10^23}$

$\\ \sf\bull\longmapsto No\:of\:moles=0.7mol$

Molar mass of CaCO3

$\\ \sf\bull\longmapsto 40u+12u+3(16u)$

$\\ \sf\bull\longmapsto 52u+48u$

$\\ \sf\bull\longmapsto 100g/mol$

Now

$\\ \sf\bull\longmapsto No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}$

$\\ \sf\bull\longmapsto Given\:mass=No\:of\:moles\times Molar\:Mass$

$\\ \sf\bull\longmapsto Given\:Mass=0.7(100)$

$\\ \sf\bull\longmapsto Given\:Mass=70g$

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3 years ago

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The enthalpy of fusion of methanol (CH3OH) is 3.16 kJ/mol. How much heat would be absorbed or released upon freezing 25.6 grams

Verdich [7]

Answer:

There is 2.52 kJ of energy released (option 4)

Explanation:

Step 1: Data given

The enthalpy of fusion of methanol (CH3OH) is 3.16 kJ/mol

Mass of methanol = 25.6 grams

Molar mass of methanol = 32.04 g/mol

Step 2: Calculate moles of methanol

Moles methanol = mass methanol / molar mass methanol

Moles methanol = 25.6 grams / 32.04 g/mol

Moles methanol = 0.799 moles

Step 3: Calculate energy transfer

Energy transfer = moles * enthalpy of fusion

Energy = 0.799 moles * 3.16 kJ/mol

Energy = 2.52 kJ released

There is 2.52 kJ of energy released

6 0

3 years ago

Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou

Ulleksa [173]

Answer:

$4.36~g~XY$

Explanation:

In this case, we can start with the reaction:

$2X + Y_2~->~2XY$

If we check the reaction, we will have 2 X and Y atoms on both sides. So, the reaction is balanced. Now, the problem give to us two amounts of reagents. Therefore, we have to find the limiting reagent. The first step then is to find the moles of each compound using the molar mass:

$3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X$

$4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2$

Now, we can divide by the coefficient of each compound (given by the balanced reaction):

$\frac{0.04~mol~X}{1}=~0.04$

$\frac{0.0875~mol~Y_2}{2}=0.04375$

The smallest value is for "X", therefore this is our limiting reagent. Now, if we use the molar ratio between "X" and "XY" we can calculate the moles of XY, so:

$0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY$

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol $Y_2$ = 48 g $Y_2$ (therefore 1 mol Y = 24 g Y). With this in mind the molar mass of XY would be 85+24 = 109 g/mol. With this in mind: