These are diseases that affect primarily the lungs or any other part of the respiratory system.
<h3>What are respiratory disorders?</h3>
The term respiratory disorders refers to those diseases that affect primarily the lungs or any other part of the respiratory system. These are the diseases that prevent a person from breathing in and out normally and prevent the lungs from playing its usual role in gaseous exchange.
The respiratory disorders could be able to make a person not to be able to breathe well and it may lead to hospitalization and death of the patient of adequate treatment is not sought from an expert who could diagnose and treat such respiratory disorders.
The common types of respiratory disorders are;
- asthma
- cystic fibrosis
- emphysema
- lung cancer
- mesothelioma
- tuberculosis
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Answer:
Explanation:
Zn + 2HCl = ZnCl₂ + H₂
A ) mole of Zn = 2.65 / 65
= .04 mol
mole of HCl required = .04 x 2 mol
.08 mol
If v be the volume required
v x 6 = .08
v = .0133 liter
= 13.3 cc
B )
volume of gas at NTP :
moles of gas obtained = .04 moles
= 22.4 x .04 liter
= .896 liter
we have to find this volume at given temperature and pressure


= 1.175 liter.
C )
.04 mole of zinc chloride will be produced
mol weight of zinc chloride
= 65 + 35.5 x 2
= 136 gm
.04 mole = 136 x .04
= 5.44 gm
Answer:
Evaporation and condensation are two processes through which matter changes from one state to another.
Explanation:
Answer: The empirical formula for the given compound is 
Explanation : Given,
Percentage of H = 18 %
Percentage of N = 82 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of H = 18 g
Mass of N = 82 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Hydrogen = 
Moles of Nitrogen = 
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 5.8 moles.
For Hydrogen = 
For Nitrogen = 
Step 3: Taking the mole ratio as their subscripts.
The ratio of H : N = 3 : 1
Hence, the empirical formula for the given compound is 
<span>Reaction of ethane combustion:
2C2H6 (g) + 7O2 (g) ----> 4CO2 (g) + 6H2O
According to the reaction, we can see that </span>C2H6 and CO2 have following stoichiometric ratio:
n(C2H6) : n(CO2) = 2 : 4
If we know the number of moles of ethane we can calculate moles of carbon dioxide:
2 x n(CO2) = 4 x n(C2H6)
n(CO2) = 2 x n(C2H6)
n(CO2) = 2 x 5.4 = 10.8 mole of CO2 <span>are produced</span>