Hello, here’s the answer to your question. Converting ammonia to nitrate, which is absorbed by plants
Pure substances are further broken down into elements and compounds. Mixture are physically combined structures that can be separated into their original components. A chemical substance is composed of one type of atom or molecule.
Answer:
Option a. 0.5 m/s
Explanation:
This graph shows a straight line, where "Y" axis would be "Position" and "X" graph would be "Time". The ecuation that would describe this straight line is Y= aX + 1 , where "a" is the slope or inclination for this graph, and would give us the speed of the object
How do we find the slope (and hence, the speed)?: if you notice this graph, you will check that:
-When X (Time) is zero, Y (Position) is 1
-When X (Time) is 2, Y (Position) is 2
With these 4 points, you can calculate the slope (which will call "m") for this graph with:
m = (Y2-Y1)/(X2-X1) so: Y2=2, Y1=1, X2=2, X1=0
Which gives us: m=1/2 (0.5), the slope or speed of the object: 0.5 m/s
Answer:
A chemical change occurs when the reactants chemical compositions have changed
Explanation:
a change in colour, change in temperature, change in smell, formation of a precipitate, or the formation of gas bubbles
Hopes this helps
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%
