The reaction creates dihydrogen, hence if it's uncontrolled it could lead to potentially dangerous amounts of gas being released at once.
Pluto is a dwarf planet, but one of the largest known members, in the Kuiper belt.
The Kuiper Belt extends between 30 AU and 55 AU from the Sun
(1 AU = 1.5 × 10^8 km = distance from Earth to Sun).
Pluto's orbit is highly elliptical. It ranges from 30 AU to 50 AU. When Pluto is closest to the Sun, it is inside the orbit of Neptune (30 AU).
Astronomers class Pluto as a <em>resonant Kuiper belt object</em> (KBO). Because it gets so close to Neptune, its orbit is in <em>resonance</em> with that of Neptune. Pluto makes two orbits for every three of Neptune.
25/2 and 96/X
CROSS MULTIPLY.
2x=2,400.
divide by 2.
x=1,200.
you take the GIVEN MASS of an element, and you put it on top, the coefficient is what it’s over. i believe this is right
I would assume so.
Given
, we can simplify the fraction to
Both would obtain the same proportions, so I don't see why putting a half cup of sugar would make things any different.
Hope this is the answer you are looking for.
Answer:
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Explanation:
tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=
mass compound
mass H
×100%
\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=
mass compound
mass C
×100%
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=
10.0g compound
2.5g H
×100%=25%
\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=
10.0g compound
7.5g C
×100%=75%
