Answer:
Explanation:
First reaction gives you the number of moles or the mass from Carbon and hydrogen
for carbon:
Analogously for hydrogen:
0.0310g
have 0.0034gH or 0.0034mol of H
In the second reaction you can obtain the amount of nitrogen as a percentage and find the mass of N in the first sample.
now
this is equivalet to 0.002mol of N
with this information you can find the mass of oxygen by matter conservation.
this is equivalent to 0.004molO
finally you divide all moles obtained between the smaller number of mole (this is mol of H)
and you can multiply by 5 to obtain:
The effect of an increase in the average kinetic energy of an object would be an increase in the temperature of the object. Temperature is directly proportional to the average kinetic energy of the molecules in the object. Hope this answers the question.
Answer:
<em>a= In scientific notation</em>
6.96×10⁵ Km
<em>b =In expanded notation</em>
0.00019 mm
Explanation:
Given data:
Radius of sun = 696000 Km
size of bacterial cell = 1.9 ×10⁻⁴ mm
Radius of sun in scientific notation = ?
Size of bacterial cell in expanded notation = ?
Solution:
Radius of sun:
696000 Km
<em>In scientific notation</em>
6.96×10⁵ Km
Size of bacterial cell:
1.9 ×10⁻⁴ mm
<em>In expanded notation</em>
1.9/ 10000 = 0.00019 mm
Answer:
the temperature of the star
Explanation:
The color of stars usually indicates the temperature of the star.
A star that is relatively cold usually shows a typical red color.
The hottest stars have a blue color.
- These star colors have been used by astronomers to determine their temperature.
- A broad spectrum between blue, the hottest color, and red the coldest is used.
- Class O stars are usually the blue colored ones
- Class M is the coldest with red color
Answer:
Explanation:
Given that:
The Half-life of 
= 
is less than that of 
Although we are not given any value about the present weight of .
So, consider the present weight in the percentage of to be y%
Then, the time elapsed to get the present weight of = 
Therefore;
here;
= Number of radioactive atoms relating to the weight of y of
Thus:
--- (1)
However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of to be = 
Then:
---- (2)
here;
= Number of radioactive atoms of relating to 3.0 a/o weight
Now, equating equation (1) and (2) together, we have:
replacing the half-life of =
( since 
)
∴
The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o
Thus, The time elapsed is
