Answer:
15.3 %
Explanation:
Step 1: Given data
- Mass of the sample (ms): 230 g
- Mass of carbon (mC); 136.6 g
- Mass of hydrogen (mH): 26.4 g
- Mass of nitrogen (mN): 31.8 g
Step 2: Calculate the mass of oxygen (mO)
The mass of the sample is equal to the sum of the masses of all the elements.
ms = mC + mH + mN + mO
mO = ms - mC - mH - mN
mO = 230 g - 136.6 g - 26.4 g - 31.8 g
mO = 35.2 g
Step 3: Calculate the mass percent of oxygen
%O = (mO / ms) × 100% = (35.2 g / 230 g) × 100% = 15.3 %
Explanation:
As per Brønsted-Lowry concept of acids and bases, chemical species which donate proton are called Brønsted-Lowry acids.
The chemical species which accept proton are called Brønsted-Lowry base.
(a) ![HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-](https://tex.z-dn.net/?f=HNO_3%20%2B%20H_2O%20%5Crightarrow%20H_3O%5E%2B%20%2B%20NO_3%5E-)
is Bronsted lowry acid and
is its conjugate base.
is Bronsted lowry base and
is its conjugate acid.
(b)
![CN^- + H_2O \rightarrow HCN + OH^-](https://tex.z-dn.net/?f=CN%5E-%20%2B%20H_2O%20%5Crightarrow%20HCN%20%2B%20OH%5E-)
is Bronsted lowry base and HCN is its conjugate acid.
is Bronsted lowry acid and
is its conjugate base.
(c)
![H_2SO_4 + Cl^- \rightarrow HCl + HSO_4^-](https://tex.z-dn.net/?f=H_2SO_4%20%2B%20Cl%5E-%20%5Crightarrow%20HCl%20%2B%20HSO_4%5E-)
is Bronsted lowry acid and
is its conjugate base.
Cl^- is Bronsted lowry base and HCl is its conjugate acid.
(d)
![HSO_4^-+OH^- \rightarrow SO_4^{2-}+H_2O](https://tex.z-dn.net/?f=HSO_4%5E-%2BOH%5E-%20%5Crightarrow%20SO_4%5E%7B2-%7D%2BH_2O)
is Bronsted lowry acid and
is its conjugate base.
OH^- is Bronsted lowry base and
is its conjugate acid.
(e)
![O_{2-}+H_2O \rightarrow 2OH^-](https://tex.z-dn.net/?f=O_%7B2-%7D%2BH_2O%20%5Crightarrow%202OH%5E-)
is Bronsted lowry base and OH- is its conjugate acid.
is Bronsted lowry acid and OH- is its conjugate base.
Unit of measurement
ex: ft, in, etc.
<h3>Solution-:</h3>
- option D
- maintains a constant volume.
#<em>o</em><em>f</em><em>f</em><em>i</em><em>c</em><em>a</em><em>i</em><em>l</em><em> </em><em>Nazo</em>
<em>ll </em><em>Radhe</em><em> Radhe</em><em> ll</em>