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Akimi4 [234]
3 years ago
5

How many grams are 5.82 x 10^24 molecules of H2S

Chemistry
1 answer:
kotykmax [81]3 years ago
0 0

As we calculate the the no. of grams of any molecules then we should know about every method or every conversion method how we concert gram into moles and moles into gram so in this conversion method we know that 1 mole of any atoms = 6.022*1023(avogadro's#)

So 5.82*1023 / 6.022*1023 = 9.66 gram of H2S

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Calculate the wavelength and frequency of an emitted photon of gamma radiation that has energy of 2.93 × 109 J/mol
shusha [124]

Answer:

Explanation: Calculate the wavelength and frequency of an emitted photon of gamma radiation that has energy of 2.93 × 109 J/mol

6 0
4 years ago
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
2.How many neutrons does an atom of tungsten (W) have when its mass number is 185?​
solniwko [45]

Answer:

110

Explanation:

3 0
3 years ago
The density of methanol is 0.787 g/ml. What is the mass in grams of 25.0 ml of methanol? Give your answer with the correct unit
Sveta_85 [38]

Answer:

19.6 g is the mass of methanol

Explanation:

Density of methanol is 0.787 g/mL.

Density means mass / volume

Methanol density = Methanol mass / Methanol volume.

Let's replace in the formula

0.787 g/mL = Methanol mass / 25 mL

0.787 g/mL . 25 mL = Methanol mass → 19.6 g

8 0
3 years ago
For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin
erastovalidia [21]

Answer : The initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1

k=7.5\times 10^{-3}M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1

\text{Rate}=3.4\times 10^{-3}Ms^{-1}

Therefore, the initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

8 0
3 years ago
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