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saw5 [17]
3 years ago
8

While in Europe, if you drive 125 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 40.0 mi/gal ? Assume that 1euro=1.26dollars.
Chemistry
1 answer:
LenKa [72]3 years ago
6 0

You would spend <em>$71.32</em> per week on gas.

<em>Step 1</em>. Convert <em>weeks to days </em>

Time = 1 wk × (7 days/1 wk) = 7 days

<em>Step 2</em>. Calculate the <em>distance</em> travelled

Distance = 7days × 125 km/day = 875 km

<em>Step 3</em>. Convert <em>kilometres to miles </em>

Distance = 875 km × (1 mi/1.609 km) = 543.8 mi

<em>Step 4</em>. Calculate the <em>gallons of gas </em>

<em>V</em> = 543.8 mi × (1 gal/40.0 mi) = 13.60 gal

<em>Step 5</em>. Convert <em>gallons to litres</em>

<em>V</em> = 13.60 US gal × (3.785 L/1 US gal) = 51.46 L

<em>Step 6</em>. Calculate the <em>cost of the gas</em>

Cost = 51.46 L × (€1.10/1 L) = €56.60

<em>Step 7</em>. Convert <em>euros to dollars</em>

Cost = €56.60 × ($1.26/€1) = <em>$71.32 </em>

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What is the empirical formula for a substance that is composed of 40.66% carbon, 8.53% Hydrogen,23.72% Nitrogen, and 27.09% Oxyg
prohojiy [21]

Answer:

THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO

Explanation:

The steps involved in calculating the empirical formula of this substance in shown in the table below:

Element                            Carbon           Hydrogen          Nitrogen         Oxygen

1. % Composition              40.66              8.53                 23.72                27.09

2. Mole ratio =

%mass/ atomic mass       40.66/12         8.53/1          23.72/14            27.09/16

                                       =  3.3883            8.53              1,6943             1.6931

3. Divide by smallest

value (0.6931)          3.3883/1.6931    8.53/1.6931    1.6943/1.6931   1.6931/1.6931

                               =      2.001                  5.038           1.0007                      1

4. Whole number ratio        2                       5                   1                               1

The empirical formula = C2H5NO

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3 years ago
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