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m_a_m_a [10]
2 years ago
15

A 13.7g sample of a compound exerts a pressure of 2.01atm in a 0.750L flask at 399K. What is the molar mass of the compound?a. 3

18 g/mol
b. 204 g/mol
c. 175 g/mol
d. 298 g/mol
Engineering
1 answer:
Katarina [22]2 years ago
4 0

Answer: Option D) 298 g/mol  is the correct answer

Explanation:

Given that;

Mass of sample m = 13.7 g

pressure P = 2.01 atm

Volume V = 0.750 L

Temperature T = 399 K

Now taking a look at the ideal gas equation

PV = nRT

we solve for n

n = PV/RT

now we substitute

n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )

= 1.5075 / 32.7579

= 0.04601 mol

we know that

molar mass of the compound = mass / moles

so

Molar Mass = 13.7 g / 0.04601 mol

= 297.7 g/mol  ≈ 298 g/mol

Therefore Option D) 298 g/mol  is the correct answer

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Answer:

Technician be says that diesel engines produce more power because they use excess air to burn feel who is correct

Explanation:

He is correct as many engines are run by diesel. It produces more power as that is how cars produce more power.

3 0
2 years ago
Coal fire burning at 1100 k delivers heat energy to a reservoir at 500 k. Find maximum efficiency.
Marizza181 [45]

Answer:

<em>55%</em>

Explanation:

hot reservoir = 1100 K

cold reservoir = 500 K

<em>This is a Carnot system</em>

For a Carnot system, maximum efficicency of the system is given as

Eff = 1 - \frac{Tc}{Th}

where Tc = temperature of cold reservoir = 500K

Th = temperature of hot reservoir = 1100 K

Eff = 1 - \frac{500}{1100}

Eff = 1 - 0.45 = 0.55 or<em> 55%</em>

7 0
2 years ago
Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a
Elena L [17]

Answer:

\mathbf{\tau_c =5.675 \ MPa}

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [\bar 101]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]

where;

[d_1\ e_1 \ f_1] = directional indices for tensile stress

[d_2 \ e_2 \ f_2] = slip direction

replacing their values;

i.e d_1 = 0 ,e_1 = 0 f_1 =  1 & d_2 = -1 , e_2 = 0 , f_2 = 1

cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]

cos \ \lambda = \dfrac{1}{\sqrt{2}}

Also, to find the angle \phi between the stress [001] & normal slip plane [111]

Then;

cos \  \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]

replacing their values;

i.e d_1 = 0 ,e_1 = 0 f_1 =  1 & d_3 = 1 , e_3 = 1 , f_3 = 1

cos \  \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]

cos \phi= \dfrac{1} {\sqrt{3} }

However, the critical resolved SS(shear stress) \mathbf{\tau_c} can be computed using the formula:

\tau_c = (\sigma )(cos  \phi )(cos \lambda)

where;

applied tensile stress \sigma = 13.9 MPa

∴

\tau_c =13.9\times (  \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})

\mathbf{\tau_c =5.675 \ MPa}

3 0
2 years ago
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Answer:

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Explanation:

  • Metal: Vanadium
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  • Molecuar weight: 50,9 g/mol

The Avogadro's Number, 6,022*10^23, is the number of atoms in one mole of any substance. To calculate the number of atoms in one cubic meter of vanadium we write:

1m^3*(100^3 cm^3/1 m^3)*(6,1 g/1 cm^3)*(1 mol/50,9g)*(6,022*10^23 atoms/1 mol)=7,217*10^28 atoms

Therefore, for vanadium we have 7,217*10^28 atoms/m^3

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