<span>C2H4O + H2O = C2H6O2
moles water = 18 g/ 18.02 g/mol=.998
the ratio is 1 : 1 : 1
moles C2H4O needed = 0.998
mass C2H4O = 0.998 mol x 44.053 g/mol = 44.00 g
We require 44.00 g of ethylene oxide is needed to react with 18 g of water.</span>
Answer:
Benzene: 37.5 Torr
Methylbenzene: 12.5 Torr
Explanation:
By Raoult's Law, each substance in a gas mixture contributes to the total pressure of the mixture proportionally to their respective mole fraction. So,
Ppartial = x*P°
Where x is the mole fraction (0.5 for each one because it's equimolar), and P° is the vapor pressure.
Benzene: Ppartial = 0.5 * 75 = 37.5 Torr
Methylbenzene: Ppartial = 0.5 * 25 = 12.5 Torr
Answer:
20.2 amu.
Explanation:
Let A represent isotope ²⁰X
Let B represent isotope ²²X
From the question given above, the following data were obtained:
For Isotope A (²⁰X):
Mass of A = 20
Abundance (A%) = 90%
For Isotope B (²²X):
Mass of B = 22
Abundance (A%) = 10%
Relative atomic mass (RAM) =?
The relative atomic mass (RAM) of the element can be obtained as follow:
RAM = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]
RAM = [(20 × 90)/100] + [(22 × 10)/100]
RAM = 18 + 2.2
RAM = 20.2 amu
Thus, relative atomic mass (RAM) of the element is 20.2 amu
Answer:
1. Chemical reactions
2.Substances
3. Properties
4. Precipitate
5. Light
6. Temperature
7. Color
8. Gas
Explanation:
During chemical reactions, new substances that possess new properties are formed. During the process of chemical reactions, evidences are found. Some of the evidences are formation of precipitate or light gel, production of gases and changes in color and temperature.
Chemical reactions start with the substances or compounds which are known as reactants. The reactants react to form new substances known as the products which possess new properties.
I only could help with 1 - 8 sorry.
55.9 kPa; Variables given = volume (V), moles (n), temperature (T)
We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:
<em>pV = nRT</em>
Solve for <em>p</em>: <em>p = nRT/V</em>
R = 8.314 kPa.L.K^(-1).mol^(-1)
<em>T</em> = (265 + 273.15) K = 538.15 K
<em>V</em> = 500.0 mL = 0.5000 L
∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa
