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Feliz [49]
3 years ago
9

While in Europe, if you drive 109 km per day, how much money would you spend on gas in one week if gas costs 1.10 euros per lite

r and your car's gas mileage is 22.0 mi/gal ? Assume that 1euro=1.26dollars.
Chemistry
1 answer:
MrRissso [65]3 years ago
5 0

Answer:

The amount is x  =  113.3 \  dollars

Explanation:

From the question we are told that

     The  distance traveled per day is  l  =  109\ km =  \frac{109}{1.609}  =  67.74 \  mi

    The  cost of one liter is  c =  1.10 \ euros/liter = 1.10  *  1.26  = 1.36 \ dollars/liter

     The car's  gas mileage is  b =  22.0 \ mi/gal

Generally the amount of distance covered in one week is evaluated as

      z =  67.74 * 7

       z = 474.18 \  mi

The  amount of gas used in one week by the car is mathematically represented as  

       k  = \frac{ z}{ b}

=>    k  = \frac{ 474.18}{ 22}

=>      k  =  22 \ gal

converting to liters

          k =  22 *  3.78541=83.28 \ liters

Thus the amount spent on gas in one week is  

          x =  k *  c

=>      x  =  83.28 *  1.36

=>      x  =  113.3 \  dollars

 

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When 91.5 g of isopropyl alcohol which has an empirical formula of C3H8O is burned in excess oxygen gas, how many grams of H2O a
Kobotan [32]

Answer:

109.7178g of H2O

Explanation:

First let us generate a balanced equation for the reaction. This is illustrated below:

2C3H8O + 9O2 —> 6CO2 + 8H2O

Next we will calculate the molar mass and masses of C3H8O and H20. This is illustrated below:

Molar Mass of C3H8O = (3x12.011) + (8x1.00794) + 15.9994 = 36.033 + 8.06352 + 15.9994 = 60.09592g/mol.

Mass of C3H8O from the balanced equation = 2 x 60.09592 = 120.19184g

Molar Mass of H2O = (2x1.00794) + 15.9994 = 2.01588 + 15.9994 = 18.01528g/mol

Mass of H2O from the balanced equation = 8 x 18.01528 = 144.12224g

From the equation,

120.19184g of C3H8O produced 144.12224g of H20.

Therefore, 91.5g of C3H8O will produce = (91.5 x 144.12224) /120.19184 = 109.7178g of H2O

7 0
3 years ago
How many molecules of N204 are in 85.0 g of N2O4?
alukav5142 [94]

Answer:

5.56 × 10^23 molecules

Explanation:

The number of molecules in a molecule can be calculated by multiplying the number of moles in that molecule by Avagadro's number (6.02 × 10^23)

Using mole = mass/molar mass

Molar mass of N2O4 = 14(2) + 16(4)

= 28 + 64

= 92g/mol

mole = 85.0/92

= 0.9239

= 0.924mol

number of molecules of N2O4 (nA) = 0.924 × 6.02 × 10^23

= 5.56 × 10^23 molecules

4 0
3 years ago
0.000431 L<br><br> Express your answer as an integer.
Molodets [167]

A whole number, not a fraction, that can be negative, positive or zero are integers. They cannot have decimal places.

Now, converting 0.000431 L to decimal an integer as:

0.000431 L = 431\times 10^{-6} L

Since, 1 L = 10^{6} \mu L

So, 431\times 10^{-6} L\times \frac{10^{6 \mu L}}{1 L} = 431 \mu L.

Hence, the integer value for 0.000431 L is 431 \mu L[.


3 0
3 years ago
How many liters of volume is one mole of gas at standard temperature and<br> pressure?
horsena [70]

22.7 liters

The molar volume of an ideal gas depends on the temperature and pressure. One mole of any ideal gas occupies 22.7 liters at 0 0C and 1 bar (STP).

Hope this helped

3 0
3 years ago
How many grams of argon would it take to fill a large light bulb with a volume of 0.745 L at STP?
EastWind [94]

Answer:

Mass = 1.33 g

Explanation:

Given data:

Mass of argon required = ?

Volume of bulb = 0.745 L

Temperature and pressure = standard

Solution:

We will calculate the number of moles of argon first.

Formula:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

By putting values,

1 atm ×0.745 L = n × 0.0821 atm.L/mol.K× 273.15 K

0.745 atm. L = n × 22.43 atm.L/mol

n = 0.745 atm. L / 22.43 atm.L/mol

n = 0.0332 mol

Mass of argon:

Mass = number of moles × molar mass

Mass = 0.0332 mol × 39.95 g/mol

Mass = 1.33 g

8 0
3 years ago
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