<span>It generally does not mean that there is double the oxygen, but in this case there is double, because the subscript number tells how many atoms of that element are in a particle. In this case, there are two of the oxygen, hence the DI-oxide verbiage, and one of the carbon. When there is only one, it's MONOxide, to indicate only one atom.</span>
Answer:
D. 15.8atm
Explanation:
Given parameters:
Initial pressure = 13atm
Initial temperature = 34°C = 34 + 273 = 307K
Final temperature = 100°C = 100 + 273 = 373K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we apply a derivation of the combined gas law taking the volume as a constant.
The expression is shown mathematically below;
= 
P and T pressure and temperature values
1 and 2 are initial and final states
Insert the parameters and solve for T₂;
= 
P₂ = 15.8atm
Answer:
A. percentage mass of iron = 5.17%
percentage mass of sand = 8.62%
percentage mass of water = 86.205%
B. (Iron + sand + water) -------> ( iron + sand) ------> sand
C. The step of separation of iron and sand
Explanation:
A. Percentage mass of the mixtures:
Total mass of mixture = (15.0 + 25.0 + 250.0) g =290.0 g
percentage mass of iron = 15/290 * 100% = 5.17%
percentage mass of sand = 25/290 * 100% = 8.62%
percentage mass of water = 250/290 * 100% = 86.205%
B. Flow chart of separation procedure
(Iron + sand + water) -------> separation by filtration using filter paper and funnel to remove water --------> ( iron + sand) -----------> separation using magnet to remove iron ------> sand
C. The step of separation of iron and sand by magnetization of iron will have the highest amount of error because during the process, some iron particles may not readily be attracted to the magnet as they may have become interlaced in-between sand grains. Also, some sand particle may also be attracted to the magnet as they are are borne on iron particles.
Answer: Decreasing the temperature inside the container will decrease the pressure of a gas inside a closed cubical container.
Explanation:
According to Gay-Lussac's Law : 'The pressure of the gas increases with increase in temperature of the gas when volume of the gas is kept constant'.
At constant volume, pressure of the gas will decrease on decreasing the temperature or vice versa.
Decreasing the temperature inside the container will decrease the pressure of a gas inside a closed cubical container.
