Answer:
For Mass N, Mass H, and Mass O, the mass is 28.0 g N, 4.0 g H, and 48.0 g respectively
Explanation:
The computation of the mass of each element is given below:
As we know that
A1 mole of ammonium nitrate i.e. 2 mol N, 4 mol H, 3 mol
Now we multiply each of above by the molar masses
For N
= 14.0 g/mol × 2
= 28.0 gN
For H
= 1.0 g/mol × 4
= 4.0 gN
ANd, for O
= 16.0 g/mol × 3
= 48.0 gN
Hence, For Mass N, Mass H, and Mass O, the mass is 28.0 g N, 4.0 g H, and 48.0 g respectively
Answer:
Explanation:
concentration unit relates moles of solute to volume of solution. ... First find the number of moles of KCl in the 25.00 mL of 0.500 M solution: ... 13) What volume of 1.25 M sulfuric acid is needed to dissolve 0.750 g of ... mL of 0.0962 M hydrochloric acid is titrated with a calcium hydroxide solution, and ...
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
Answer:
15g
Explanation:
The equation for the reaction is given below:
CaO + SO2 —> CaSO3
Molar Mass of CaO = 40 + 16 = 56g/mol
Molar Mass of CaSO3 = 40 + 32 + (16x3) = 40 + 32 + 48 = 120g/mol
From the equation,
56g of CaO produced 120g of CaSO3.
Therefore, 7g of CaO will produce = (7 x 120)/56 = 15g of CaSO3
Therefore, 15g of CaSO3 is produce
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.
