Answer:
Explanation:
Florine has the highest electronegativity and it thus pulls the element of 
groups in the COOH group towards itself, making it relatively easy to loose the proton of the carboxyl group.
F O
↑ ||
F← C ← C ← O - H , Here three Florine withdraw
↓
F
F O
↑ ||
F← C ← C - O H , Here two Florine withdraw
↓
H
F O
| ||
H - C - C - OH , Here one Florine withdraw
↓
H
H O
| ||
H - C - C - OH , Here no Florine withdraw
|
H
A quantitative observation is an observation obtained from using instruments or tools such as balances, rulers, beakers and thermometers. The results here are measurable. The opposite is called the quantitative observations where the senses are used to see a result.
Answer:
- <em>As the temperature of a sample of matter is increased, the average kinetic energy of the particles in the sample </em><u>increase</u><em>.</em>
Explanation:
The <em>temperature</em> of a substance is the measure of the <em>average kinetic energy </em>of its partilces.
The temperature, i.e. how hot or cold is a substance, is the result of the collisions of the particles (atoms or molecules) of matter.
The kinetic theory of gases states that, if the temperature is the same, the average kinetic energy of any gas is the same, regardless the gas and other conditions.
This equation expresses it:
Where Avg KE is the average kinetic energy, R is the universal constant of gases, N is Avogadro's constnat, and T is the temperature measure in absolute scale (Kelvin).
As you see, in that equation Avg KE is propotional to T, which means that as the temperature is increased, the average kinetic energy increases.
Answer:
7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100 M solution.
Explanation:
First of all the molecular weight of Na2SO4 is 142.08 gram.Now we all know that if the molecular weight of a compound is dissolved in 1000ml or 1 litee of water then the strength of that solution becomes 1 M.
According to the given question we have to prepare 0.100 M solution
1000 ml of solution contain 142.08×0.1= 14.208 gram Na2SO4
1 ml of solution contain 14.208÷1000= 0.014 gram
0.5L or 500ml of solution contain 0.014×500= 7gram Na2SO4.
So it can be stated that 7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100M solution.
<span><span>N2</span><span>O5</span></span>
Explanation!
When given %, assume you have 100 g of the substance. Find moles, divide by lowest count. In this case you'll end up with
<span><span>25.92 g N<span>14.01 g N/mol N</span></span>=1.850 mol N</span>
<span><span>74.07 g O<span>16.00 g O/mol O</span></span>=4.629 mol O</span>
The ratio between these is <span>2.502 mol O/mol N</span>, which corresponds closely with <span><span>N2</span><span>O5</span></span>.
