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Readme [11.4K]
3 years ago
12

CO is a primary pollutant.

Chemistry
2 answers:
Anuta_ua [19.1K]3 years ago
7 0

True

Carbon monoxide is a primary pollutant which no odor results from incomplete combustion of fuel. The man sources are gasoline and burning of biomass.

Depending on the source of emission, pollutants can be classified into two groups that is primary and secondary pollutants.

A primary pollutant is emitted in the atmosphere directly from a source. It can be either natural sch as volcanic eruptions, sandstorms or man-made that is due to industrial and vehicle emissions.  Examples of primary pollutants are nitrogen oxides, carbon monoxide and particulate matter.

Secondary pollutant is due to interactions between primary and secondary pollutants. These can be chemical or physical interactions. Examples are photo-chemical oxidants and secondary particulate matter.

Therefore, carbon monoxide CO is a primary pollutant.





Sonja [21]3 years ago
4 0
TRUEEEEEEEEEEEEEEEEE
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1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

8 0
3 years ago
Exactly 5000 mL of air at 223K is warmed and has a new volume of 8.36 liters. What is the new temperature?
34kurt

Answer:

The new temperature is 373 K

Explanation:

Step 1: Data given

Volume air = 5000 mL = 5.0 L

Temperature = 223K

New volume = 8.36 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒V1 = the initial volume = 5.0 L

⇒T1 = the initial temperature = 223 K

⇒V2 = the new volume = 8.36 L

⇒T2 = the new temperature

5.0/223 = 8.36 /T2

T2 = 373 K

The new temperature is 373 K

7 0
3 years ago
How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?
Montano1993 [528]

Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

so you have (144.3x1.4/100) g glucose= 2.0202 grams

6 0
3 years ago
Read 2 more answers
A nitrogen atom has five valence electrons. The Lewis structure shows how a nitrogen atom bonds with three hydrogen atoms to for
nordsb [41]
It has too be 6 or 4

3 0
3 years ago
What are the differences between simple, compound, and electron microscopes? WILL GET BRAINIEST!
timurjin [86]
I think the answer is= simple/uses surrounding light source and is restricted in magnification.
compound uses an electrical light source but is restricted in magnification also. an electron microscope has electrical magnification and light source so you can see smaller cells when dyed the rt. color..plse double check, good luck..
5 0
3 years ago
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