Answer:
5.72 moles of nitrogen oxide are formed when 2.86 moles of nitrogen react.
Explanation:
Given data:
Moles of NO formed = ?
Moles of N₂ reacts = 2.86 mol
Solution:
Chemical equation:
N₂ + O₂ → 2NO
Now we will compare the moles of nitrogen with nitrogen oxide from balance chemical equation.
N₂ : NO
1 : 2
2.86 : 2/1×2.86 = 5.72 mol
So 5.72 moles of nitrogen oxide are formed when 2.86 moles of nitrogen react.
Set up a proportion os V1/n1 = V2/n2. V is the volume, n is the amount in MOLES, not grams. Convert the CO_2 to moles, then solve and find that the amount of N_2 should be the same amount of moles. Then use the molar mass of N_2 (28.02 grams/mole) to convert that amount of moles into grams. That's your answer.
generally melting point of mixture is always different from melting point of pure compound
lets say you have a two compounds called A and B
and you have a unknown compound called X which can be either A or B
now your target is to find unknown is really A or B right?
take some amount of X and mix it with A and B separately
out of these two mixtures one will be pure compound (because we initially assume that unknown can be either A or B)
suppose X + A is same melting with A then X is A
X +B will be less than the melting point of A or
suppose X+B is same melting point like B then X is B
X+A will be less than melting point of B
Answer:
0.4 M
Explanation:
M=moles/Liters
Convert grams to moles (446/223.37=1.99668711)
Convert 5000 mL to liters (5000/1000)
Plug in the values and solve
M=2/5
M=0.4
The correct answer for the question that is being presented above is this one: "A. The accepted model of the atom was changed.<span>" </span>J J Thomson discovered the electrons and performed experiment using the cathode ray tube
Here are the following choices:
<span>A. The accepted model of the atom was changed.
B. The accepted model of the atom was supported.
C. Cathode ray tubes were no longer used in experiments due to poor results.
D. Cathode ray tubes became the only instrument of use in the study of atoms</span>
