What can you infer about arid climates from their name?

Determine the formula weights of each of the following compounds.Part A) Nitrous oxide, N2O, known as laughing gas and used as a

densk [106]

Answer:

See explanation

Explanation:

a) Nitrous oxide (N2O) has a molar mass of 44.014 amu. It has 2 nitrogen atoms each with a mass of 14.007 amu and 1 oxygen atom with a mass of 16.0 amu.

Percentage nitrogen = (2*14.007 amu/ 44.014 amu ) *100% = 63.6%

Percentage oxygen = (16 amu/44.014 amu) * 100% = 36.4 %

63.6% + 36.4% = 100%

b) Benzoic acid (C7H6O2) has a molar mass of 122.13 amu. It has 6 hydrogen atoms each with a mass of 1.01 amu; it has 7 carbon atoms each with a mass of 12.01amu and 2 oxygen atoms with a mass of 16.0 amu.

Percentage hydrogen = (6*1.01 amu / 122.13 amu)*100% = 4.96%

Percentage carbon = (7*12.01 amu/ 122.13 amu)*100% = 68.8%

Percentage oxygen = (2*16 amu/ 122.13 amu) *100% = 26.2%

c) Magnesium hydroxide (Mg(OH)2) has a molecular mass of 58.32 amu. It has 2 hydrogen atoms each with a mass of 1.01 amu; it has 1 magnesium atom with a mass of 24.3 amu and two oxygen atoms with a mass of 16.0 amu.

Percentage hydrogen = (2*1.01 amu/ 58.32 amu) *100% = 3.46 %

Percentage magnesium = (24.3 amu/58.32 amu)*100% = 41.7%

Percentage oxygen = (2*16 amu/58.32 amu)*100% = 54.9%

d) Urea CO(NH2)2 has a molecular mass of 60.064 amu. It has 2 Nitrogen atoms each with a mass of 14.007 amu, 4 hydrogen atoms each with a mass of 1.01 amu,1 carbon atom with a mass of 12.01 amu and 1 oxygen atom with a mass of 16.0 amu.

Percentage nitrogen = (2*14.007 amu/ 60.064amu)*100% = 46.6%

Percentage hydrogen = (4*1.01 amu/60.064amu)*100% = 6.72%

Percentage carbon = (12.01 amu/60.064amu)*100% = 20.0%

Percentage oxygen = (16 amu/60.064amu)*100% = 26.6%

e) Osopentyl acetate (C7H14O2) has a molecular mass of 130.2 amu. It has 14 hydrogen atoms each with a mass of 1.01 amu,7 carbon atoms each with a mass of 12.01 amu and 2 oxygen atom with a mass of 16.0 amu.

Percentage hydrogen = (14*1.01 amu/130.2 amu)*100% = 10.8%

Percentage carbon = (7*12.01 amu/130.2 amu)*100% = 64.6%

Percentage oxygen =(2*16 amu/130.2 amu)*100% = 24.6%

7 0

3 years ago

Please someone help me write a Conclusion about climate change

Tatiana [17]

Answer:

Here u go

Explanation:

climate change...climate change is when the weather changes drastically and the adaptations change so that some animals environment are no longer suitable for them

I tried to hel

4 0

3 years ago

2NO(g) + O2 (g) --> 2NO2

Lorico [155]

Take 46g / 46 g/mol = 1 mol NO2 produced
1 mol * 1/2 = 0.5 mol O2 needed (1/2 ratio between NO2 and O2)

4 0

2 years ago

Pls help with this

Natalka [10]

Hirlkrrkkrkrmmrrmmejekkeowoow

3 0

2 years ago

The reaction: 2 CO(l) + O2(g) ⇄2 CO2(g) with a H = −25.0 kJ/mol, is at equilibrium. Which of the following lists three ways thi

elena-s [515]

Answer:

Option B and D both have 3 ways to shift the reaction to the right

Explanation:

The reaction 2 CO(l) + O2(g) ⇄2 CO2(g) is exothermic because ΔH = -25 kJ < 0 This means heat will be released

Therefore, increasing the temperature will shift the equilibrium to the left, while decreasing the temperature will shift the equilibrium to the right.

A. remove CO2, lower pressure and remove CO

⇒ Lowering the pressure will shift the equilibrium to the side with most moles of gas : On the left side there are 3 moles, on the right side 2 moles. By lowering the pressure, the equilibrium will shift to the left.

B remove CO2, raise pressure and add CO

⇒ By raising the pressure, the equilibrium will shift to the side with the lesser amount of moles of gas. This is the right side.

⇒ Remove CO2: the equilibrium will shift to the side of CO2 ,so the reaction will shift toward products to replace the product removed. (The right side).

⇒ Add CO: the equilibrium will shift to the side of CO2, so the reaction will shift toward the products side to reduce the added CO. (The right side).

C raise temperature, lower pressure and remove O2

⇒ Increasing the temperature will shift the equilibrium to the left

D add O2, raise pressure and lower temperature

⇒ decreasing the temperature will shift the equilibrium to the right

⇒ By raising the pressure, the equilibrium will shift to the side with the lesser amount of moles of gas. This is the right side.

⇒ Add O2: the equilibrium will shift to the side of CO2, so the reaction will shift toward the products side to reduce the added O2. (The right side).

E remove CO2, increase volume and lower temperature

⇒ decreasing the temperature will shift the equilibrium to the right

⇒ Remove CO2: the equilibrium will shift to the side of CO2 ,so the reaction will shift toward products to replace the product removed. (The right side).

⇒ Increase volume : with a pressure decrease due to an increase in volume, the side with more moles is more favorable. The equilibrium will shift to the left side.

8 0

3 years ago