17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C. Details about how to calculate mass can be found below.
<h3>How to calculate mass?</h3>
The mass of a given gas can be calculated by multiplying the number of moles of the substance by its molar mass.
However, the number of moles of the gas must be calculated first as follows:
PV = nRT
Where;
- P = pressure = 1.0796941atm
- V = volume = 12.3L
- n = number of moles
- T = temperature = 288.4K
- R = gas law constant = 0.0821 Latm/molK
1.079 × 12.3 = n × 0.0821 × 288.4
13.27 = 23.68n
n = 13.27/23.68
n = 0.56mol
Mass = 0.56 × 32
mass of oxygen gas = 17.93g
Therefore, 17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C.
Learn more about mass at: brainly.com/question/19694949
Answer:
Dark clothing
Explanation:
Dark clothing like black absorbs/locks in heat. Since exercise is to burn and tonify our body from unwanted fat it is important to sweat. This is where heat comes in and helps you sweat
Answer:
a) Aqueous LiBr = Hydrogen Gas
b) Aqueous AgBr = solid Ag
c) Molten LiBr = solid Li
c) Molten AgBr = Solid Ag
Explanation:
a) Aqueous LiBr
This sample produces Hydrogen gas, because the H+ (conteined in the water) has a reduction potential higher than the Li+ from the salt. Therefore the hydrogen cation will reduce instead of the lithium one and form the gas.
b) Aqueous AgBr
This sample produces Solid Ag, because the Ag+ has a reduction potential higher than the H+ from the water. Therefore the silver cation will reduce instead of the hydrogen one and form the solid.
c) Molten LiBr
In a molten binary salt like LiBr there is only one cation present in the cathod. In this case the Li+, so it will reduce and form solid Li.
c) Molten AgBr
The same as the item above: there is only one cation present in the cathod. In this case the Ag+, so it will reduce and form solid Ag.
No, some species didn’t have bones and things of the such that could become fossilized.
8.5L H2O (1 mol H2O/22.4L)=.37946 mol H2O
.3796 mol H2O(1 mol CH4/2 mol H2O)=.18973 mol CH4
.18973 mol CH4(22.4L/1mol CH4)=4.25L CH4 gas
