5.6L of O2 means we have 0.25 moles of O2.
As, 1 mole has 6.023*10^23 molecules,
0.25 moles of O2 will have 0.25*6.023*10^23 molecules=1.50575*10^23 molecules
and as 1 molecule of O2 has 2 atoms, so, 1.50575*10^23 molecules will have 2*1.50575*10^23 atoms=3.0115*10^23 atoms of O.
Answer:
A) 7.9 x 10⁶ inches
B) 1004 g
C) 2.8 x 10³ inches/ min
D) 1.2 x 10⁻⁴ mm
Explanation:
A) Since 39.37 inches = 1 m, you can convert meters to inches by multiplying by the conversion factor (39.37 inches / 1 m).
Notice that if 39.37 inches = 1 m then 39.37 inches / 1 m = 1. That means that when you multiply by a conversion factor, you are only changing units since it is the same as multiplying by 1 :
2.0 x 10⁵ m * (39.37 inches / 1 m) = 7.9 x 10⁶ inches
B) Conversion factors : (2.205 pounds / 1 kg) and (453.59 g / 1 pound), because 2.205 pounds = 1 kg and 1 pound = 453.59 g. Then:
1.004 kg * ( 2.205 pounds / 1 kg) * ( 453.59 g / 1 pound) = 1004 g
C) Conversion factor: (39.37 inches / 1 m) and (60 s / 1 min)
1.2 m/s * (39.37 inches / 1 m) * ( 60 s / 1 min) = 2.8 x 10³ inches/ min
D)Converison factor ( 1 mm / 1 x 10⁶ nm):
120 nm (1 mm / 1 x 10⁶ nm) = 1.2 x 10⁻⁴ mm
Complete Question:
This diagram shows a marble with a mass of 3.8 grams (g) that was placed into 10 milliliters (mL) of water. Using the formula V M D = , what is the density of the marble?
(See attachment for full diagram)
Answer:
1.27 g/cm³
Explanation:
First, find the volume of rock:
Volume of rock = volume of water after rock was placed - volume of water before rock was placed
Volume of rock = 13 - 10 = 3ml
Density of rock = grams of rock per 1 cm³
Note: 1 ml = 1 cm³
Let x represent amount of rock per 1 cm³
Thus,
3.8g = 3 cm³
x = 1 cm³
Cross multiply
1*3.8 = 3*x
3.8 = 3x
3.8/3 = 3x/3
1.27 = x
Density of rock = 1.27 g/cm³
The shape of the organism is a huge factor, as it could fact for a different creature and/or organism every time. For example, a antelope may have a different form or structure of let’s say a spider, as this is the significant factor of ones shape, the first observation.
Secondly, the colors. As different organism come in different areas of reflections of light, it can easily be identified with this certain color and the way that the light reflects on it. This is another form of visual appearance, and important factor in identification.
Lastly, the pattern. As located on the back, the spikes/thread-looking things sticking out plays a huge factor in finding the organism name and/or identifying it. Every organism has a different pattern, humans being a huge one, all different
Observational studies of organisms is significant to our lives, and it’s greatly helped progress our sciences and society.
