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otez555 [7]
3 years ago
14

Which development was a direct result of the

Chemistry
2 answers:
xxTIMURxx [149]3 years ago
7 0
<span>the development that became a direct result of thebaby boom that followed World War II would be: increase in suburbanization Baby boomers makes the urban city became a lot more crowded and often caused heavy traffic every single day. Because if this, many people decided to live in a suburban area while commuting to work in the city at the same time</span>
Keith_Richards [23]3 years ago
6 0
<span>Many countries experienced the baby boom after the World War II ended. The baby boom resulted to the increase in suburbanization which is number 2. The more people/babies born, the more houses are needed. The countries who experienced the baby boom were not large enough to cater the population so mass developments happened.</span>
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Limiting reactants would appreciate the help
Vanyuwa [196]

Answer:

Explanation:

The Limiting Reactant is that reactant which when consumed in a reaction stops the reaction. The other reactants will be in excess and typically considered non-reactive.

To identify the limiting reactant ...

- write and balance the reaction of interest. Express it in standard form. That is, standard form of a reaction is when the coefficients of the balanced equation are in their lowest whole number values. Also, remember that the standard equation is 'assumed' to be at STP conditions (0°C & 1atm).

- convert all given reactant values to moles

- divide each reactant mole value by the related coefficient of the the balanced standard equation. The smaller value is the limiting reactant. The remaining reactants will be in excess.  

Your Problem:

Given:        3Ba  +  N₂  => Ba₃N₂

               22.6g    4.2g        ?

moles Ba => 22.6g/137.34g/mol = 0.165 mole Ba

moles N₂ =>    4.2g/14.007g/mol= 0.150 mole N₂

Part A: Determining the Limited Reactant

  • Divide each mole value by respective coefficient ... smallest value is Limiting  Reactant.

Barium => 0.165/3 = 0.055  <=> (Limiting Reactant)

Nitrogen => 0.15/1 = 0.15

  • Barium is the smaller result and is therefore the limiting reactant. This works for ALL limiting reactant type problems. However, be sure to use the mole values calculated first (Ba = 0.165mol & N₂ = 0.150mol) when doing ratio calculations.

Part B: Max (theoretical) amount of Ba₃N₂ produced:

<em>Note: The product yield amounts are based upon the given 'moles' of limiting reactant, NOT the results of the 'divide by respective coefficient' step used to ID the limiting reactant.     </em>

                   3Ba        +          N₂         =>     Ba₃N₂    (3:1 rxn ratio for Ba:Ba₃N₂)

moles      0.165mole        0.150mole         1/3(0.165)mole = 0.055mole Ba₃N₂

                                                                    = 0.055mol(440g/mol) Ba₃N₂

                                                                    = 24.2 grams Ba₃N₂ (as based

                                                                     upon Barium as Limiting Reactant)

Part C: Excess N₂ remaining after reaction stops:

From balanced standard reaction, the reaction ratio for Ba:N₂ is 3moles:1mole. That is, for the moles of Ba consumed, 1/3(moles of Ba) =  moles of N₂ used.

moles of N₂ used = 1/3(0.165)mole = 0.055mole N₂ used  

∴ the amount of N₂ remaining in excess = 0.150mole (given) - 0.055mole (used) = 0.095mole N₂ remaining in excess.

mass N₂ remaining = 0.095mole x 28g/mole = 2.66 grams N₂ remaining in excess.

                   

5 0
3 years ago
What is the oxidation number for iodine in Mg(IO3)2 ?
navik [9.2K]
The oxidation number of iodine is 5 in Mg(IO3)2 which can be calculated as 
   Mg(IO3)2
   MgI2O6
As we know that
Mg has +2
O has -2
So,
   (+2) + 2I + 6 (-2)=0
   2 + 2I - 12 =0
   10+ 2I =0
    10 = 2I
     I =5

7 0
3 years ago
Read 2 more answers
Calculate Ecell at 80 ºC for a voltaic cell based on the following redox reaction: H2(g, 1.25 atm) + 2AgCl(s) → 2Ag(s) + 2H+(aq,
iris [78.8K]

Answer:

Ecell = +0.25V

Explanation:

the half-cell reactions for a voltanic cell

cathode(reduction): 2H⁺(aq) + 2e⁻ ------- H₂(g)

anode(oxidation): 2AgCl(s) ------- 2Ag⁺(aq) + 2Cl⁻ + 2e⁻

we have the standard cell potential E⁺cell = 0.18V at 80C respectively

Q = [H⁺]/[Cl⁻]

sub for [H+] = 0.10M and [Cl-] = 1.5M

Q= 0.1M/1.5M

Q = 0.067

Ecell = E⁺cell - \frac{0.059}{n} logQ

= 0.18 - \frac{0.056}{1} log 0.067

0.18- 0.059(-1.174)

Ecell = +0.25V

6 0
3 years ago
Describe how you can figure out how many grams of a COMPOUND if you know how many moles there
____ [38]
Divide the mass of the compound in grams by the molar mass you just calculated. The answer is the number of moles of that mass of compound. For example, 25 grams of water equals 25/18.016 or 1.39 moles.
4 0
2 years ago
HELP ASAP PLEASE!!!
Dmitry [639]

Answer:

A. Electrolyte

Explanation:

Concentrated sulfuric acid has a density of 1.84 g/millimeter. When you dilute this with water to 5.20 M, you then have a density of 1.30 g/millimeter, which then can be used as a lead storage for batteries in automobiles. (Got help to answer this at Www.wyzant.com

3 0
3 years ago
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