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Vitek1552 [10]
3 years ago
5

A sample of N2O3(g) has a pressure of 0.046 atm . The temperature (in K) is then doubled and the N2O3 undergoes complete decompo

sition to NO2(g) and NO(g). Find the total pressure of the mixture of gases assuming constant volume and no additional temperature change. Enter your answer numerically, in terms of atm.
Chemistry
1 answer:
iogann1982 [59]3 years ago
5 0

Answer:

0.184 atm

Explanation:

The ideal gas equation is:

PV = nRT

Where<em> P</em> is the pressure, <em>V</em> is the volume, <em>n</em> is the number of moles, <em>R</em> the constant of the gases, and <em>T</em> the temperature.

So, the sample of N₂O₃ will only have its temperature doubled, with the same volume and the same number of moles. Temperature and pressure are directly related, so if one increases the other also increases, then the pressure must double to 0.092 atm.

The decomposition occurs:

N₂O₃(g) ⇄ NO₂(g) + NO(g)

So, 1 mol of N₂O₃ will produce 2 moles of the products (1 of each), the <em>n </em>will double. The volume and the temperature are now constants, and the pressure is directly proportional to the number of moles, so the pressure will double to 0.184 atm.

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Suppose the price of gasoline increases 10% and quantity of gasoline demanded in Orlando drops 5% per day. Demand for gasoline i
grin007 [14]

Answer:

The correct answer is;

Demand for gasoline in Orlando is price inelastic.

Explanation:

The elasticity is the degree of response to a change in price or quantity supplied to the the quantity demanded. An elastic demand responds positively to change in price, while an inelastic demand means that when there is a price increase, the quantity demanded remains the same and where there is a drop in price the quantity demanded remains constant.

If a small change in price results in a large change in demand then the good is said to be price elastic

In the question the price increases by 10% while the quantity demanded drops 5 % daily. Therefore it is price inelastic

5 0
3 years ago
T/F___ At the eutectic composition, an alloy can solidify at a constant temperature.___ For effective dispersion strengthening,
azamat

Answer:

  • TRUE
  • FALSE
  • TRUE
  • FALSE
  • FALSE
  • TRUE

Explanation:

  • At the eutectic composition, an alloy can solidify at a constant temperature : TRUE . this is because at eutectic composition the type of reaction that takes place there is invariant reaction in its thermal equilibrium
  • For effective dispersion strengthening, the dispersed phase should be needle-like, as opposed to round : FALSE. because the rounded shape will not cause a crack.
  • Intermetallic compounds are usually hard and brittle : TRUE. because Intermetallic compounds prevents dislocation movements and this makes them brittle and hard
  • For the effective dispersion and strengthening, the dispersed phase should be continuous : FALSE. this is because the dispersed precipitate must be small and not continuous
  • Stoichiometric intermetallic compounds exist over a range of compositions : FALSE
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4 0
3 years ago
Copper is formed when aluminum reacts with copper (II) sulfate in a single-replacement reaction. How many moles of copper can be
serg [7]

Answer:

The answer to your question is the limiting reactant is CuSO₄ and 0.975 moles of Cu were obtained

Explanation:

moles of Copper = ?

mass of Aluminum = 29 g

mass of CuSO₄ = 156 g

Limiting reactant = ?

Balanced Chemical reaction

                  3 CuSO₄   +  2 Al   ⇒   3 Cu   +  Al₂(SO₄)₃

Calculate the moles of reactants

CuSO₄ = 64 + 32 + (16 x 4) = 160g

Al = 27 g

                160 g of CuSO₄  ----------------- 1 mol

                156 g                   -----------------  x

                      x = (156 x 1) / 160

                      x = 0.975 moles

               27 g of Al -------------------------- 1 mol

               29 g of Al -------------------------- x

                x = (29 x 1)/27

                x = 1.07 moles

Calculate proportions to find the limiting reactant

Theoretical     3 moles CuSO₄/2 moles Al = 1.5 moles

Experimental  0.975 moles CuSO₄/1.07 moles = 0.91

The experimental proportion was lower than the theoretical proportion that means that the limiting reactant is CuSO₄.      

                    3 moles of CuSO₄ ------------------ 3 moles of Cu

                 0.975 moles of CuSO₄ ---------------  x

                         x = (0.975 x 3)/3

                        x = 0.975 moles of Cu were obtained.        

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