The percentage of hydrogen in C7H18 is calculated as follows:
[18/(12*7+1*8)]*100=18%
The amount of hydrogen in 5.2moles is given by:(18/100)*5.2=0.94moles
Answer:
a= <em>In scientific notation</em>
6.96000×10⁵ Km
b =<em>In expanded notation</em>
0.00019 mm
Explanation:
Given data:
Radius of sun = 696000 Km
size of bacterial cell = 1.9 ×10⁻⁴ mm
Radius of sun in scientific notation = ?
Size of bacterial cell in expanded notation = ?
Solution:
Scientific notation is the way to express the large value in short form.
The number in scientific notation have two parts.
. The digits (decimal point will place after first digit)
× 10 ( the power which put the decimal point where it should be)
for example the number 6324.4 in scientific notation will be written as = 6.3244 × 10³
Radius of sun:
696000 Km
<em>In scientific notation</em>
6.96000 × 10⁵ Km
The expanded notation is standard notation of writing the numerical values which is normal way. The numbers are written as they are, without the power of 10.
Size of bacterial cell:
1.9 ×10⁻⁴ mm
<em>In expanded notation</em>
1.9/ 10000 = 0.00019 mm
Answer is: because weak acids do not dissociate completely.
The strength of an Arrhenius
acid determines percentage of ionization of acid and the number of H⁺ ions formed. <span>
Strong acids completely ionize in water and give large amount ofhydrogen ions (H</span>⁺), so we use only one arrow, because reaction goes in one direction and there no molecules of acid in solution.
For example hydrochloric acid: HCl(aq) → H⁺(aq) + Cl⁻(aq).
<span>
Weak acid partially ionize in water
and give only a few hydrogen ions (H</span>⁺), in the solution there molecules of acid and ions.
For example cyanide acid: HCN(aq) ⇄ H⁺(aq)
+ CN⁻(aq).
Compound: a thing that is composed of two or more separate elements, basically it is a mixture
Mixture: a substance made by mixing other substances together
so go from there. I don't want to cheat by giving answers, so I hope this is guiding and helpful. please mark brainliest if it is!
Answer:
This is an oxidation-reduction (redox) reaction:
2 Ni0 - 4 e- → 2 NiII
(oxidation)
2 O0 + 4 e- → 2 O-II
(reduction)
Ni is a reducing agent, O2 is an oxidizing agent.
