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1 year ago

15

A particular car has a gas mileage of 24.5 miles per gallon. If the cost of gas is $4.25 per gallon, and the car travels at a co

nstant speed of 35 miles per hour, what is the cost to drive for 300 minutes?

1 answer:

tester [92]1 year ago

3 0

Answer:

$30.39

Explanation:

alot of math

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Like many other organic solvents we will use this semester, ethanol is flammable and caution needs to be exercised when heating

laiz [17]

Answer:

See explanation

Explanation:

A flammable solvent refers to a solvent that catches fire easily. The precautions to be taken when working with flammable solvents are;

1) heat the solvent at a low to medium hot plate setting.

2) if you need to boil the solvent, use a condenser rather than a flask or beaker without a cover.

3) make sure that the hotplate is larger than the vessel containing the mixture that is being heated.

4) do not use strong oxidizing agents

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2 years ago

Oxygen gas, generated by the reaction 2KClO3(s)---2KCl(s)+3O2(g), is collected over water at 27•C in 3.72L vassel at a total pre

Julli [10]

Answer:

moles = 0.093 moles

Explanation:

In this case, we know that this reaction is taking plave in a vessel that has a 730 torr of total pressure.

The total pressure is a value obtained by:

Pt = Pwater + PO2

We need to know the pressure of O2, because then, with stoichiometry, we can calculate the moles of KClO3

The pressure of oxygen is:

PO2 = 730 - 26 = 704 Torr

Now, this pressure is in Torr, and we need to convert it to Atm, so:

704 Torr / 760 Torr = 0.9263 atm

Now, let's use the ideal gas equation:

PV = nRT

With this expression, we will calculate the moles of O2, and then, the moles of KClO3:

n = PV/RT

R = 0.082 L atm /K mol

P = 0.9263 atm

V = 3.72 L

T = 27 + 273 = 300 K

Replacing the data:

n = 0.9263 * 3.72 / 300 * 0.082

n = 0.14 moles

Finally, by stoichiometry, we know that 2 moles of KClO3 produces 3 moles of O2, so:

moles of KClO3 = 0.14 * 2/3 = 0.093 moles of KClO3

6 0

1 year ago

Butane (c4h10) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. given δh°f[c4h10(g)] = –124.7

vagabundo [1.1K]

The value of Δ H butane (g) = -124.7 kJ/mol

The value of Δ H CO2 (g) = -393.5 kJ/mol

The value of Δ H H2O (g) = -241.8 kJ/mol

Mass of butane, m = 8.30 gm

Molar mass of butane is 58 gm/mol

Consider the reaction,

C₄H₁₀ + 6.5 O₂ = 4CO₂ + 5H₂O

Calculating the value of Δ H° rxn:

ΔH°rxn = ∑nH° f (products) - ∑nH° f (reactants)

Substituting the values we get,

Δ H° rxn = 4 (-393.5) + 5 (-241.8) - (-124.7)

= -1574 -1209 + 124.7

= -2783 - 124.7

= -2658.3 kJ/mol

Now, calculate the number of moles of butane in 8.30 gm.

Number of moles = mass/molar mass

= 8.30 / 58

= 0.143 moles

Thus, the total energy released in the reaction is,

Q = number of moles × ΔH° rxn

= 0.143 × (2658.3)

= 380.14 kJ

Hence, the total heat released in the reaction is 380.14 kJ.

6 0

2 years ago

a sample of an oxide of iron was reduced to iron by heating with hydrogen. the mass of iron obtained was 4.35g and mass of water

Hitman42 [59]

Answer:

FeO(s) + H2(g)→ Fe(s) + H2O(g)

Explanation:

Moles of Iron will be = 4.35 g/55.845 g/mol

= 0.0786 moles

Moles of water = 1.86 g/18 g/mol

= 0.103 moles

The mole ratio of Iron to water

= 0.0786 : 0.103

= 1 : 1.3

= 1 : 1

Therefore, the equation for the reaction is;

FeO(s) + H2(g)→ Fe(s) + H2O(g)

5 0

1 year ago

The volume of a gas is 27.5 mL at 22.0°C and 0.974 atm. What will the volume be at 15.0°C and 0.993 atm? Use Ideal Gas Law (PV =

Scorpion4ik [409]

Answer:

26.3 mL

Explanation:

Step 1:

Obtaining an appropriate gas law from the ideal gas equation.

This is illustrated below:

From the ideal gas equation:

PV = nRT

Divide both side by T

PV/T = nR

At this stage, we'll assume the number of mole (n) to be constant.

Note: R is the gas constant.

PV/T = constant.

We can thus, write the above equation as:

P1V1/T1 = P2V2/T2

The above equation is called the general gas equation.

Step 2:

Data obtained from the question. This includes the following:

Initial volume (V1) = 27.5 mL

Initial temperature (T1) = 22.0°C = 22.0°C + 273 = 295K

Initial pressure (P1) = 0.974 atm.

Final temperature (T2) = 15.0°C = 15.0°C + 273 = 288K

Final pressure (P2) = 0.993 atm

Final volume (V2) =..?

Step 3:

Determination of the final volume of the gas using the general gas equation obtained. This is illustrated below:

P1V1 /T1 = P2V2/T2

0.974 x 27.5/295 = 0.993 x V2/288

Cross multiply to express in linear.

295x0.993xV2 = 0.974x27.5x288

Divide both side by 295 x 0.993

V2 = (0.974x27.5x288)/(295x0.993)

V2 = 26.3 mL

Therefore, the new volume of the gas is 26.3 mL

4 0

2 years ago