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2 years ago

Explain whether nitrogen atoms will form bonds with other atoms

Chemistry

1 answer:

iren2701 [21]2 years ago

3 0

The closer the atoms are the more the nitrogen will be come bigger or the oppisite

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Given that a commercial airliner has a range a range of 9,500 miles using 265 tons of JP-4 fuel. Estimate the range of the aircr

Blizzard [7]

Here is the full question.

Given that a commercial airliner has a range of 9,500 miles using 265 tonnes of JP-4 fuel. Estimate the range of the aircraft burning either gaseous or liquid hydrogen at the same mass of hydrogen as JP4.

The dry mass of the aircraft is 800 tonnes. The fuel properties are as follows:

Fuel Density (kg/m³) Heating value (kJ/kg)

JP4 800.0 45000

H₂ (gaseous, S.T.P) 0.0824 120900

H₂ (liquid, 1 atm) 70.8 120900

Answer:

25,514 mi

Explanation:

Let first calculate the value of initial mass of the aircraft $m_1$ when burning JP4 fuel by using the expression:

$m_1 =m_2 + m_{JP4}$

where;

$m_2 = 800t$

$m_{JP4} = 265t$

$m_1 = 800t+265t$

$m_1=1065 t$

We need to employ the use of the range of the airliner which can be expressed by the formula;

$s = \eta__0}(\frac{L}{\delta g})In(\frac{m_1}{m_2})(\frac{Q_R}{g})$

where;

$\eta__0}$ = overall efficiency

L = lift

$\delta g$ = drag force

$m_1 =$ initial mass of the vehicle

$m_2$ = is the final mass of the airliner after the burning of fuel

$Q_R$ = is the heat of the reaction of the fuel burning.

g = gravitational acceleration

Rearranging the above equation; we have:

$\eta__0}(\frac{L}{\delta g}) = \frac{ s}{In(\frac{m_1}{m_2})({Q_R})}$

$\eta__0}(\frac{L}{\delta g}) = \frac{9500 mi}{In(\frac{1065t}{800t})({45000 kJ/kg})}$

$\eta__0}(\frac{L}{\delta g}) = \frac{0.2111}{In(1.33)}$

$\eta__0}(\frac{L}{\delta g}) = 0.74 mi.kg/kJ$

To estimate the range of the aircraft burning either gaseous or liquid hydrogen at the same mass of hydrogen as JP4; we have:

$s = \eta__0}(\frac{L}{\delta g})In(\frac{m_1}{m_2})(\frac{Q_R}{g})$

$s= (0.74mi.kg/kJ)(120900kJ/kg)In(\frac{1065t}{800t})$

$s= (89466mi)In(\frac{1065t}{800t})$

$s = (89466mi)In(1.33)$

s = 25,513.82 mi

s ≅ 25,514 mi

Thus, the range of the aircraft when burning either gaseous or liquid hydrogen at the same mass of hydrogen as JP4 is 25,514 mi

6 0

3 years ago

Find the number of moles of water that can be formed if you have 226 mol of hydrogen gas and 108 mol of oxygen gas.

Anit [1.1K]

Answer is: there are 216 moles od water.
Chemical reaction: 2H₂ + O₂ → 2H₂O.
n(H₂) = 226 mol.
n(O₂) = 108 mol; limiting reactant.
From chemical reaction: n(O₂) : n(H₂O) = 1 : 2.
n(H₂O) = 2 · n(O₂).
n(H₂O) = 2 · 108 mol.
n(H₂O) = 216 mol.
n - amount of substance.

7 0

3 years ago

How many moles are in 1.52908x1024 molecules of Potassium phosphate (K3(PO4))?

lara31 [8.8K]

Answer:

<h2>2.54 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$