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Alex73 [517]
3 years ago
8

If your car gets 30.3 mi/gal, how many gallons of gasoline would you use if you drove 506.3 miles? 1 mile = 5280 ft (exactly) In

clude one decimal place in your answer.
Chemistry
2 answers:
mr_godi [17]3 years ago
7 0

Answer : The number of gallons of gasoline needed are 16.7 gallons

Explanation :

As we are given that the car traveled 30.3 mi/gal. Now we have to determine the number of gallons of gasoline needed in 506.03 miles.

As, 30.3 miles car traveled in = 1 gallon

and, 1 miles car traveled in = \frac{1\text{ mile}}{30.3\text{mile}}\times 1 \text{ gallon}

So, 506.3 miles car traveled in = \frac{506.3\text{ mile}}{30.3\text{mile}}\times 1 \text{ gallon}

                                                    = 16.7 gallons

Therefore, the number of gallons of gasoline needed are 16.7 gallons

Salsk061 [2.6K]3 years ago
4 0

Answer: 16.7 gallon

Explanation:

Given: The car can drive 30.3 miles when 1 gallon of gasoline is used.

Distance covered = 506.3 miles

Thus for 30.3 miles, the amount of gasoline used= 1 gallon

For 506.3 miles, the amount of gasoline used=\frac{1}{30.3}\times 506.3=16.7gallon

Thus the amount of gasoline used is 16.7 gallons.


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Calculate moles of oxygen atoms in 0.68mol of KMnO4
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Alright, so that means we have 0.68 mol of the compound


For each 1 mol of the compound, we have 4*1 oxygens (because there are four oxygens in the formula)
Therefore for each 0.68 mol of the compound, we have 4*0.68 moles of oxygen!
6 0
2 years ago
Beryllium oxide, Beo, is an electrical insulator. How
egoroff_w [7]

Answer:

There are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.

Explanation:

We can calculate the number of moles (η) of BeO as follows:

\eta = \frac{m}{M}

Where:

m: is the mass = 250 g

M: is the molar mass = 25.0116 g/mol

Hence, the number of moles is:

\eta = \frac{250 g}{25.0116 g/mol} = 10.0 moles

Therefore, there are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.  

I hope it helps you!

3 0
3 years ago
URGENT !! A substance has 55.80% carbon, 7.04% Hydrogen, and 37.16% Oxygen. What is it's empirical and molecular formula if it h
Ede4ka [16]
<h3><u>Answer;</u></h3>

Empirical formula = C₂H₃O

Molecular formula = C₁₄H₂₁O₇

<h3><u>Explanation</u>;</h3>

Empirical formula

Moles of;

Carbon = 55.8 /12 = 4.65 moles

Hydrogen = 7.04/ 1 = 7.04 moles

Oxygen  = 37.16/ 16 = 2.3225 moles

We then get the mole ratio;

4.65/2.3225 = 2.0

7.04/2.3225 = 3.0

2.3225/2.3225 = 1.0

Therefore;

The empirical formula = <u>C₂H₃O</u>

Molecular formula;

(C2H3O)n = 301.35 g

(12 ×2 + 3× 1 + 16×1)n = 301.35

43n = 301.35

  n = 7

Therefore;

Molecular formula = (C2H3O)7

                             <u> = C₁₄H₂₁O₇</u>

6 0
3 years ago
The initial explanation for a phenomenon prior to experimentation is called a:
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The explanation for a phenomenon is a hypothesis something happened that you sure that was going to happen
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3 years ago
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