What is the density (g/mL) of a sample of mineral oil if 250 mL has a mass of 0.23 kg?

The Society of Automotive Engineers has established an accepted numerical scale to measure the viscosity of motor oil. For examp

tatuchka [14]

Because its molecules can slide around each other, a liquid has the ability to flow. The resistance to such flow is called the viscosity. For organic liquids, as the chain increases the viscosity increases as well due to the bonding that is present. Therefore, the ranking should be as follows:

CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH3

5 0

2 years ago

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A person throws a ball up into the air, and the ball falls back toward Earth. At which point would the kinetic energy be the low

dybincka [34]

It would be d bc when it’s at its lowest point it means not that much energy and that would be its lowest point which is D

6 0

2 years ago

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Exactly 500 grams of ice are melted at a temperature of 32°f. (lice = 333 j/g.) calculate the change in entropy (in j/k). (give

denpristay [2]

Entropy Change is calculated by (Energy transferred) / (Temperature in kelvin)
deltaS = Q / T

Q = (mass)(latent heat of fusion)
Q = m(hfusion)
Q = (500g)(333J/g) = 166,500J

T(K) = 32 + 273.15 = 305.15K
deltaS = 166,500J / 305.15K
deltaS = 545.63 J/K

3 0

2 years ago

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0

2 years ago

A species has the following MO configuration: (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(σ2p)2(π2p)2. This substance is:_______.

Goshia [24]

Answer : The correct option is, (a) paramagnetic with two unpaired electrons.

Explanation :

According to the molecular orbital theory, the general molecular orbital configuration will be,

$(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),(\sigma_{2p_z}),[(\pi_{2p_x})=(\pi_{2p_y})],[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)$

As there are 14 electrons present in the given configuration.

The molecular orbital configuration of molecule will be,

$(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,(\sigma_{2p_z})^2,[(\pi_{2p_x})^1=(\pi_{2p_y})^1],[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0$

The number of unpaired electron in the given configuration is, 2. So, this is paramagnetic. That means, more the number of unpaired electrons, more paramagnetic.

Hence, the correct option is, (a) paramagnetic with two unpaired electrons.

3 0

2 years ago