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Murljashka [212]
3 years ago
14

If your front lawn is 18.0 feet wide and 20.0 feet long. And each square foot of lawn accumulates 1450 new snow flakes every min

ute, how much snow (in kilograms) accumulates on your lawn per hour ? Assume an avarage snow flake has a mass of 1.90 mg
Chemistry
1 answer:
creativ13 [48]3 years ago
6 0
To solve this problem, we begin by first calculating the area of the front lawn. The length and width of the lawn was given and the area of a rectangle is given by the formula: Area = length x width. Thus, the area of the front lawn can be obtained by multiplying 18 ft by 20 ft, wherein we get 360 ft^2 as the area. 

Second, the problem indicated that each square foot of lawn accumulates 1450 new snow flakes per minute. This can be translated into the expression 1450 snow flakes/ (minute·ft^2). In this way, we can convert it to units of mass (kg). Afterwards, we simply need to multiply it to the area of the lawn and convert minute to hour. The following expression is then used:

1450 snow flakes/ (minute·ft^2) x 1.90 mg/snow flake x 1 g/1000 mg x 1kg/1000 g x 360 ft^2 x 60 minutes/hour = 59.508 kg snow flake/hour

It is then calculated that 59.508 kg of snow flake accumulates in the lawn every hour. 
 
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Answer:

P_1=206.7lbmol/h\\\\P_2=413.3lbmol/h

Explanation:

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In this case, for the separation of the benzene-toluene mixture, we can use the following mole balances including the given mole fraction at each stream per species:

B:0.36*620=0.98P_1+0.05P_2\\\\T:0.64*620=0.02P_1+0.95P_2

That can be solved by using a solver for P1 (benzene-rich flow) and P2 (toluene-rich flow):

P_1=206.7lbmol/h\\\\P_2=413.3lbmol/h

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The production of heat, light, or smoke is observed.

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Answer:

\large \boxed{\text{197.4 g}}

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:     28.01               17.03

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m/g:                          240.0

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\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}

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\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}

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\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}

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