The partial pressure of oxygen in a sample of air increases if the temperature is increased.
Answer: Option 1
<u>Explanation:
</u>
According to Guy-Lussac's law, at constant volume, pressure exhibited by the gas molecules will be directly proportional to the temperature of the gas molecules. It is also known that pressure of mixture of gas molecules is the sum of partial pressure of each gas molecule in the mixture.
If the temperature increases, the partial pressure and the pressure of the mixture of gas also tend to increase. As it can be seen that at higher altitudes, the low temperature leads to the decrease in oxygen's partial pressure in the air.
So, it can also be concluded that temperature increases the oxygen's partial pressure in air increases.
The large piece of jewelry that has a mass of 132.6 g and when is submerged in a graduated cylinder that initially contains 48.6 ml water and the volume increases to 61.2 ml once the piece of jewelry is submerged, has a density of: 10.523 g/ml
To solve this problem the formulas and the procedures that we have to use are:
Where:
- d= density
- m= mass
- v= volume
- v(f) = final volume
- v(i) = initial volume
Information about the problem:
- m = 132.6 g
- v(i) = 48.6 ml
- v(f) = 61.2 ml
- v = ?
- d =?
Applying the volume formula we get:
v = v(f)-v(i)
v = 61.2 ml - 48.6 ml
v = 12.6 ml
Applying the density formula we get:
d = m/v
d = 132.6 g/12.6 ml
d = 10.523 g/ml
<h3>What is density?</h3>
It is a physical quantity that expresses the ratio of the body mass to the volume it occupies.
Learn more about density in: brainly.com/question/1354972
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The most condensed state of matter is A. Solid Matter
Answer:
Explanation:
We can calculate the volume of the oxygen molecule as the radius of oxygen molecule is given as 2×10⁻¹⁰m.
We know that volume=4/3×πr³
volume =4/3×π(2.0×10⁻¹⁰m)³
volume=33.40×10⁻³⁰m³
Volume of oxygen molecule=33.40×10⁻³⁰m³
we know the ideal gas equation as:
PV=nRT
k=R/Na
R=k×Na
PV=n×k×Na×T
n×Na=N
PV=Nkt
p is pressure of gas
v is volume of gas
T is temperature of gas
N is numbetr of molecules
Na is avagadros number
k is boltzmann constant =1.38×10⁻²³J/K
R is real gas constant
So to calculate pressure using the formula;
PV=NkT
P=NkT/V
Since there is only one molecule of oxygen so N=1
P=[1×1.38×10⁻²³J/K×300]/[33.40×10⁻³⁰m³
p=12.39×10⁷Pascal
Answer:
21.8 grams.
Explanation:
Molar mass data from a modern periodic table:
How many moles of MgO will be produced if Mg is the limiting reactant?
Number of moles of Mg:
.
The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.
How many moles of MgO will be produced if O₂ is the limiting reactant?
Number of moles of O₂:
.
The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO.
of MgO will be produced if O₂ is in excess.
How many moles of MgO will be produced?
0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.
What's the mass of 0.541284 moles of MgO?
Formula mass of MgO:
.
Mass of 0.541284 moles of MgO:
.