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}](https://tex.z-dn.net/?f=C_%7B6%7DH_%7B8%7DO_%7B6%7D)
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](https://tex.z-dn.net/?f=C_%7B6%7D%3D12.0107g%2A6%3D72.08g%5C%5C%5C%5CH_%7B8%7D%3D1.008g%2A8%3D8.064g%5C%5C%5C%5CO_%7B6%7D%3D15.999g%2A6%3D95.994g%5C%5C%5C%5CC_%7B6%7DH_%7B8%7DO_%7B6%7D%3D72.08g%2B8.064g%2B95.994g%3D176.138g)
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)](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B72.08g%7D%7B176.138g%7D%2A100%3D0.409%2A100%3D40.92%28percent%29%5C%5C%5C%5CH%3D%5Cfrac%7B8.064g%7D%7B176.138g%7D%2A100%3D4.58%28percent%29%5C%5C%5C%5CO%3D%5Cfrac%7B95.994%7D%7B176.138g%7D%2A100%3D54.49%28percent%29)
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}](https://tex.z-dn.net/?f=C%3D12.0107%5Cfrac%7Bg%7D%7Bmol%7D%5C%5C%5C%5CH%3D1.008%5Cfrac%7Bg%7D%7Bmol%7D%5C%5C%5C%5CO%3D15.999%5Cfrac%7Bg%7D%7Bmol%7D%7Bmol%7D)
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](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B40.92g%7D%7B12.010%5Cfrac%7Bg%7D%7Bmol%7D%7D%3D3.41mol%5C%5C%5C%5CH%3D%5Cfrac%7B4.58g%7D%7B1.008%5Cfrac%7Bg%7D%7Bmol%7D%7D%3D4.54mol%5C%5C%5C%5CO%3D%5Cfrac%7B54.49g%7D%7B15.999%5Cfrac%7Bg%7D%7Bmol%7D%7D%3D3.40mol)
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!
The decomposition time : 7.69 min ≈ 7.7 min
<h3>Further explanation</h3>
Given
rate constant : 0.029/min
a concentration of 0.050 mol L to a concentration of 0.040 mol L
Required
the decomposition time
Solution
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time
For first-order reaction :
[A]=[Ao]e^(-kt)
or
ln[A]=-kt+ln(A0)
Input the value :
ln(0.040)=-(0.029)t+ln(0.050)
-3.219 = -0.029t -2.996
-0.223 =-0.029t
t=7.69 minutes
Answer:
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice
Explanation:
Copper is a metal with an atomic number of 29. This metal is soft and reddish in color which explains why it is very malleable(beaten to form various shapes without breaking).
All metals are good conductors of electricity including copper which is also a metal. Metals generally are insoluble in water. Copper also has a high melting point which is a characteristic of metals due to their giant structure and metallic lattice which makes it difficult to be broken down.
Mg + Cl₂ = MgCl₂M(Mg) = 24г/моль m 1 моль Mg = 24 г.По условию задачи дано 12г. Mg Количество вещества n(Mg) =12÷24=0,5 мольРассуждаем: по уравнению реакции с 1 моль магния реагирует 1 моль хлора, следовательно с 0,5 моль будет реагировать 0,5 моль хлора.1 моль хлора при н.у. занимает объем 22,4л. , тогда 0,5 моль хлора займет:0,5х22,4л.= 11,2л. Ответ: Для взаимодействии 12 г. магния потребуется 11,2 л. хлора.
2.49 x 10^46 is the answer