The maximum CO2, concentration for natural CO2, cycles from over 800,000 years?

How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?

Y_Kistochka [10]

There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

$C_{6}H_{8}O_{6}$

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

$C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g$

To know the percent of each element, we need to to the following:

$C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)$

So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

4.58 grams of H

54.49 grams of O

We know the molecular masses of each element:

$C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}$

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

$C=\frac{40.92g}{12.010\frac{g}{mol}}=3.41mol\\\\H=\frac{4.58g}{1.008\frac{g}{mol}}=4.54mol\\\\O=\frac{54.49g}{15.999\frac{g}{mol}}=3.40mol$

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

5 0

3 years ago

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

3 years ago

How many atoms of both elements are there in this molecule: 3N205; O 3 Nitrogen and 15 Oxygen 6 Nitrogen and 5 Oxygen KD 2 Nitro

abruzzese [7]

Answer:

Explanation:

oxygen is a 15 and nitrogen science chemistry i'm guessing

6 0

3 years ago

The compound ncl3 is nitrogen trichloride, but alcl3 is simply aluminum chloride. why

AveGali [126]

The answer to this question would be:
NCl3 is a molecular compound (two or more nonmetals), and therefore in its name prefixes indicate the number of each type of atom. so NCl3 is nitrogen trichloride.
The compound AlCl3 is an ionic compound (metal and nonmetal), and therefore does not require prefixes. so AlCl3 is aluminum chloride.

Both of nitrogen and chlorine is nonmetal, but aluminum is metal. Metal with nonmetal will make an ionic compound that doesn't need prefixes.

5 0

3 years ago

Read 2 more answers

Determine the number of 3s electrons in Na .

Scorpion4ik [409]

Answer:

Explanation:The atomic number of sodium is 11. That is, the number of electrons in sodium is 11. Therefore, a sodium atom will have two electrons in the first shell, eight in the 2nd orbit, and an electron in the 3rd shell.

3 0

1 year ago

The maximum CO2, concentration for natural CO2, cycles from over 800,000 years?​

The maximum CO2, concentration for natural CO2, cycles from over 800,000 years?