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Oliga [24]
3 years ago
14

Which air pollutant contributes to asthma?

Chemistry
1 answer:
Mnenie [13.5K]3 years ago
8 0
The correct answer is the second option. Particulate matter is the air pollutant known to affect people with asthma. It also can cause other illnesses like cough, chest discomfort and a burning sensation in the lungs.
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In the following combustion reaction, what is the mole ratio of ethane (C2H6) to carbon dioxide? 2 C2H6 + 7 O2(g) 4 CO2 + 6 H2O
Gwar [14]
The answer you are looking four is 4/2 or 4CO2/2C2H6
4 0
3 years ago
Read 2 more answers
9 Cardiac disease is a major problem in the United States."
amid [387]

Answer:

D It affects the heart

Explanation:

5 0
3 years ago
Given the equilibrium constants for the following two reactions at a 298K:NiO(s) + H2(g) ⇌ Ni(s) + H2O(g) Kc=40NiO(s) +CO(g) ⇌ N
shusha [124]

Answer:

The value  is  K_C =  \frac{40}{600}

Explanation:

From the question

   The equation given is  

            NiO(s) + H2(g) ⇌ Ni(s) + H2O(g) Kc=40          (1)

            NiO(s) +CO(g) ⇌ Ni(s) +CO2(g) Kc=600         (2)

Generally the reverse of the second equation as shown below

            Ni(s) +CO2(g) ⇌  NiO(s) + CO(g)                 (3)

The equilibrium constant becomes    K_c  ' =  \frac{1}{600}  

 Now  adding  1  and  3 we obtain

      NiO(s) + H2(g)+ Ni(s) +CO2(g) ⇌ Ni(s) + H2O(g)+NiO(s) + CO(g)        

               CO2(g) + H2(g) ⇌ CO(g) + H2O(g)

Hence the equilibrium constant for the resulting equation is mathematically evaluated as

           K_C =  K_c' *  K_c

=>       K_C =  \frac{1}{600}  *  40

=>         K_C =  \frac{40}{600}

       

6 0
3 years ago
How many grams of nitric acid HNO₃, are required to neutralize (completely react with) 4.30 grams of Ca(OH)2 according to the ac
Brrunno [24]

Answer:

7.32g of HNO3 are required.

Explanation:

1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.

From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.

2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:

• starting with the 4.30 grams of Ca(OH)2.

,

• using the molar mass of Ca(OH)2 (74g/mol).

,

• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .

,

• using the molar mass of HNO3 (63.02g/mol).

4.30g\text{ Ca\lparen OH\rparen}_2*\frac{1\text{ mol Ca\lparen OH\rparen}_2}{74g\text{ Ca\lparen OH\rparen}_2}*\frac{2\text{ moles HNO}_3}{1\text{ mole Ca\lparen OH\rparen}_2}*\frac{63.02g\text{ HNO}_3}{1\text{ mole HNO}_3}=7.32g\text{ HNO}_3

So, 7.32g of HNO3 are required.

4 0
2 years ago
PLZ HELP ME WITH MY WORK​
Delvig [45]

Answer:

neutrons should be added with the protons

4 0
3 years ago
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