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Elza [17]
3 years ago
10

Prior to hosting an international soccer match, the local soccer club needs to replace the artifical turf on their field with gr

ass turf. The grass turf will cost $ 9.75 per square meter. If the field is 0.102 km by 0.069 km, how much will it cost the club to add the grass turf to their field?
Chemistry
1 answer:
GuDViN [60]3 years ago
4 0

Answer:

It will cost $68,620.5 to the club to add the grass turf to their field.

Explanation:

length of the field = l = 0.102 km = 0.102 × 1000 m

( 1km = 1000 m)

Width of the field w = 0.069 km = 0.069 × 1000 m

Area of the field , A= l × w

A=0.102\times 1000 m\times 0.069\times 1000 m=7,038 m^2

Cost of grass turfing = 9.75 \$/m^2

Cost of grass turfing on field of 7,038 m^2 :

=7,038 m^2\times 9.75 \$/m^2=\$68,620.5

It will cost $68,620.5 to the club to add the grass turf to their field.

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3 years ago
Describe how you would set up an experiment to test the rela-tionship between completion of assigned homework and the fi-nal gra
ValentinkaMS [17]

To start this test, you need to identify the variables it presents. As you may already know, there are independent and dependent variables. Independent variables are those that act on a factor, influencing it to generate a result. In the case of this experiment, the independent variable is the completion of the homework. The dependent variable, in turn, is the factor that receives the influence of the independent variable, in this experiment this variable is the final grade you received in the course.

After that you must select a number of students, give them their homework and ask each student to complete a percentage of that amount. An example of this could be that you select 11 students and ask the first to complete 0% of the homework, the second student must complete 10%, the third 20% and so on, and the 11th student must complete 100% of the homework.

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7 0
3 years ago
A solution is prepared from 4.5701 g of magnesium chloride and 43.238 g of water. The vapor pressure of water above this solutio
balandron [24]

Answer:

i = 2.483

Explanation:

The vapour pressure lowering formula is:

Pₐ = Xₐ×P⁰ₐ <em>(1)</em>

For electrolytes:

Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ

Where:

Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)

4.5701g of MgCl₂ are:

4.5701g ₓ (1mol / 95.211g) = 0.048000 moles

43.238g of water are:

43.238g ₓ (1mol / 18.015g) = 2.400 moles

Replacing in (1):

0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm

0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)

2.4mol + i*0.048mol = 2.4mol / 0.9527

2.4mol + i*0.048mol = 2.5192mol

i*0.048mol = 2.5192mol - 2.4mol

i = 0.1192mol / 0.048mol

<em>i = 2.483</em>

<em />

I hope it helps!

4 0
3 years ago
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