Answer:
Density = 4.191 gm/L
Explanation:
Given:
Molar mass = 93.89 g/mol
Volume(Missing) = 22.4 L (Approx)
Find:
Density at STP
Computation:
Density = Mass/Volume
Density = 93.89 / 22.4
Density = 4.191 gm/L
Answer:
4.02
Explanation:
The mass ratios will be given by dividing the mass of O₂ into the mass of N₂.
So lets do our calculations:
First Compound:
53.3 g O₂ / 46.7 g N₂ = 1.14
Second Compound:
82.0 g O₂ / 17.9 g N₂ = 4.58
Ratio = 4.58 / 1.14 = 4.02
This result for all practical purposes is a whole number, and it is telling us that there are 4 times as many oxygen atoms in the second coumpound as in the first compound. This is so because the ratio we just calculated is also the ratio in mol atoms:
Ratio = [ mass O₂ / MW O2/ mass N₂/ MW N₂] 2nd compound / [mass O₂ / MW O2/ mass N₂/ MW N₂ !st compound]
and the molecular weights cancel each other.
The only N and O compounds that follow this ratio are N₂O₄ and N₂O, and this question could be made in a multiple choice to match formulas.
The answer is
<span>The density (D) is quotient of mass (m) and
volume (V):
</span>
The unit is g/cm³
It is given:
m = 1.62 kg = 1620 g
V = 205 mL = 205 cm³
D = ?
Thus:
The density of the goblet is 7.90 g/cm³.
<span>In order to calculate the surface are of the sheet of paper in square centimeters you must first convert the dimensions of the paper to centimeters. You can do this by multiplying 8.5 * 2.54 and 6.5 * 2.54. The paper measures 21.59cm by 16.51cm. To find the surface area you multiply the dimensions of the paper to equal 356.4509cm2</span>
Answer:
No, Stephanie is incorrect. Formation of petroleum cannot take place under the presence of oxygen.
Explanation:
Since, the petroleum is fossil product. Fossil fuel are formed under high pressure and temperature with absence of oxygen for longer period. so the way she is performing is completely incorrect. With the presence of oxygen in no way petroleum will be formed. The temperature and pressure should be in different combination for the formation of the petroleum. Along with the layers of sediments to maintain the pressure is required.
