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EastWind [94]
4 years ago
8

A military jet cruising at an altitude of 12.0km and speed of 1300./kmh burns fuel at the rate of 46.2/Lmin. How would you calcu

late the amount of fuel the jet consumes on a 1150.km mission? Set the math up. But don't do any of it. Just leave your answer as a math expression. Also, be sure your answer includes all the correct unit symbols. fuel=consumed1330kmh⋅
Chemistry
1 answer:
ira [324]4 years ago
3 0

Answer:

2,452.12 L of fuel the jet consumes on a 1150 km mission.

Explanation:

Speed of the jet = 1300 km/h

1 hour = 60 minutes

1300 km/h=\frac{1300 km}{60 min}=21.667 km/min

Rate at which jet consumes fuel = 46.2 L/min

Distance covered by jet by consuming 1 liter fuel or mileage = R

R = \frac{21.667 km/min}{46.2 L/min}=0.4690 km/L

The amount of fuel the jet consumes on a 1150 km mission will = V

Amount of fuel = \frac{Distance}{Mileage}

V=\frac{1150 km}{0.4690 km/L}=2,452.12 L

2,452.12 L of fuel the jet consumes on a 1150 km mission.

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lisabon 2012 [21]

Explanation:

XCl _{2(aq)} + 2AgNO _{3(aq)}→X(NO _{3}) _{2(aq)}   +2 AgCl _{(s)}

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3 years ago
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What is the percent of hydrogen by mass in CH4O?<br> (H:1.00 amu, C: 12.01 amu, O: 16.00 amu)
cricket20 [7]

The mass percent of hydrogen in CH₄O is 12.5%.

<h3>What is the mass percent?</h3>

Mass percent is the mass of the element divided by the mass of the compound or solute.

  • Step 1: Calculate the mass of the compound.

mCH₄O = 1 mC + 4 mH + 1 mO = 1 (12.01 amu) + 4 (1.00 amu) + 1 (16.00 amu) = 32.01 amu

  • Step 2: Calculate the mass of hydrogen in the compound.

mH in mCH₄O = 4 mH = 4 (1.00 amu) = 4.00 amu

  • Step 3: Calculate the mass percent of hydrogen in the compound.

%H = (mH in mCH₄O / mCH₄O) × 100%

%H = 4.00 amu / 32.01 amu × 100% = 12.5%

The mass percent of hydrogen in CH₄O is 12.5%.

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2 years ago
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It’s A



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3 years ago
If 1.16 L of water is initially at 24.2 ∘C, what will its temperature be after absorption of 9.4×10−2 kWh of heat?
vitfil [10]

Answer:

The temperature will be 93.92 °C

Explanation:

To explain this we will use following equation: also  Q = ∆U + W known as the NON-FLOW ENERGY EQUATION (N.F.E.E.)

With Q = heat added to the system

with ∆U  = change in internal energy

⇒∆U = ( m )( Cv )( T2 - T1 )

With W = work done by the system

⇒For this situation W = 0 because there isn't work done

So we get: ∆U = ( m )( Cv )( T2 - T1 ) = Q

To find the temperature, we have to isolate T in the equation:

(T2-T1) = Q / (m)(Cv)

⇒ Since we know that m = density * volume we can calculate the mass of water.

mass = 1000g/L * 1.16 L = 1160g

Cv = heat capacity ⇒ water has a  heat capacity of 4.184 J/g °C

We know the absorption of heat is 9.4x 10^-2 kWh but to know how many joule this is we should convert ( 1 joule = 3.6 x 10^6 kWh)

⇒Q = ( 0.094 kWh ) ( 3.6 x 10^6 J / kWh ) = 0.3384 x 10^6 J

For the temperature we get then: T2 -T1 = Q / (m)(Cv)

T2 - T1 =  0.3384 x 10^6 J / (( 1160g)*(4.184 J/g °C)) = 69.72 ° C

T2 = ( T2 - T1 ) + T1   ⇒ 69.72 + 24.2 = 93.92 °C

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3 years ago
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antiseptic1488 [7]

Answer:

B

Explanation:

i hope this help but if it didn't tell me.

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