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kolbaska11 [484]
3 years ago
8

Gas is $2.50/gal at Station A and 85 cents/L at Station B. A) Convert the cost of gas at Station B to $/gal. B) At which station

would you buy gas?
Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
8 0

Answer:

85 cents/L is equal to 3.2176$/gallon.

gas wiuld be heaper to buy from the station A.

Explanation:

As 85 cents/L is more than $2.5/gallon therefore buying gas from station A would be cheaper than the other one.

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When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H
algol [13]

Answer: The empirical formula is CH_2 and molecular formula is C_4H_8

Explanation:

We are given:

Mass of CO_2 = 18.95 g

Mass of H_2O= 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, =\frac{12}{44}\times 18.59=5.07g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, =\frac{2}{18}\times 7.759=0.862g of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.422}{0.422}=1

For H =\frac{0.862}{0.422}=2

The ratio of C : H = 1: 2

Hence the empirical formula is CH_2.

The empirical weight of CH_2 = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4

The molecular formula will be=4\times CH_2=C_4H_8

6 0
2 years ago
Suppose a 10.0 mL sample of an unknown
mote1985 [20]

The concentration of HCl is equal to 2.54mol/L.

<h3>Mole calculation</h3>

To solve this question, one must use the molarity calculation, which corresponds to the following expression:

                                               M = \frac{mol}{v}

Thus, to find the molarity of the sample, the following calculations must be performed:

V_f = 10ml + 625ml = > 635ml

                                              \frac{0.004mol}{xmol} =\frac{1000ml}{635ml}

                                                 x = 0.00254 mol

So, 0.00254 moles were added per 10ml, so we can do:

                                              \frac{0.00254mol}{xmol}= \frac{10ml}{1000ml}  \\x = 2.54mol/L

So, the concentration of HCl is equal to 2.54mol/L.

Learn more about mole calculation in: brainly.com/question/2845237

6 0
2 years ago
A gas with a vapor density greater than that of air, would be most effectively displaced out off a vessel by?
AURORKA [14]

A gas with a vapor density greater than that of air, would be most effectively displaced out off a vessel by ventilation.

The two following principles determine the type of ventilation: Considering the impact of the contaminant's vapour density and either positive or negative pressure is applied.

Consider a vertical tank that is filled with methane gas. Methane would leak out if we opened the top hatch since its vapour density is far lower than that of air. A second opening could be built at the bottom to greatly increase the process' efficiency.

A faster atmospheric turnover would follow from air being pulled in via the bottom while the methane was vented out the top. The rate of natural ventilation will increase with the difference in vapour density. Numerous gases that require ventilation are either present in fairly low concentrations or have vapor densities close to one.

3 0
1 year ago
16.
mixas84 [53]

Answer:

Explanation:

When going from top to bottom in any group of the periodic table, the atomic radius always tends to increase

8 0
2 years ago
6.0 g of copper was heated from 20 degree c to 90 degree c . How much energy was used to heat cu?
Darina [25.2K]
Copper heat capacity would be <span>0.385J/C*gram which means it needs 0.385 Joule of energy to increase 1 gram of copper temperature by 1 Celcius. The calculation would be:
energy= heat capacity *mass * temperature difference
energy= </span>0.385J/C*gram * 6g * (90-20)
<span>energy= 161.7J
 </span>
4 0
3 years ago
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