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sattari [20]
3 years ago
9

While in Europe, if you drive 107km per day, how much money would you spend on gas in one week if gas costs 1.10 euros per liter

and your car's gas mileage is 39.0mi/gal ? Assume that 1euro=1.26dollars.
Chemistry
2 answers:
Mumz [18]3 years ago
7 0
That's a lot of conversions...
107*7 = 749km in one week
749/1.6 = 468.125 miles
468.125/39 = <span>12.0032051282 gallons of fuel needed
</span>1 gallon = 4.54609 litres
1.1(Cost per litre)*4.54609(Number of litres per gallon)*12.0032051282(Number of gallons needed) = <span>60.0244158814 euros cost per week
 </span>60.0244158814 * 1.26 = <span>75.6307640105 dollar cost per week
</span>$75.63
UkoKoshka [18]3 years ago
7 0

Answer:

$ 62.578  

Explanation:

Conversion factors:

1 gal = 3.785 L

1.609km = 1 mi

$1.26 = 1 euro

1 week = 7 days

Converting the car’s mileage from mi/gal to km/L

39 mi/gal x 1 gal/3.785L x 1.609km/1 mi = 16.578 km/L

Finding the L required in a day

107 km x 1L/16.578 km = 6.45 L  

Finding the euro spent in day

1.10 euro/L x 6.45 L = 7.095 euro/ day

Finding the euro spent in a week and converting it to dollars

7.095 euro/ day x 7 days x $1.26/ 1 euro = $ 62.578  

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This is much more simple than it sounds... 

s has 1 orbital; p = 3 orbitals; d= 5 orbitals; f= 7 orbitals. (you just need to memorize this) 
A maximum of 2 electrons can occupy each orbital. 

<span>The number of orbitals that each atom has is based on the number of electrons it has and by consequence it's position on the periodic table. </span>

The orbitals occur in sequence. Whereby electrons fill first from the lowest energy level (1s) outwards to the highest. 
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4 years ago
What is a mole and its properties​
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7 0
3 years ago
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10. A sample of gas occupies 4.00 L at 80.0 kPa at 303 K. How many moles of gas are in the sample? If the sample weighs 20.50 gr
Lunna [17]

Answer:

Moles: n=0.127mol

Molar mass: M=200.61 g/mol

Explanation:

Hello,

In this case, we can use the ideal gas equation to compute the moles of the gas sample as shown below:

PV=nRT\\\\n=\frac{PV}{RT}

Thus, we should use the pressure in atm as follows:

n=\frac{80.0kPa*\frac{0.00986923atm}{1kPa}*4.00L}{0.082\frac{atm*L}{mol*K}*303K}\\ \\n=0.127mol

Moreover, the molar mass is obtained by dividing the given mass by the obtained moles:

M=\frac{m}{n}=\frac{25.50g}{0.127mol}\\  \\M=200.61 g/mol

Best regards.

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Answer:

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Answer:

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