PLZ LOOK AT ALL THE QUESTIONS. IF YOU GET THEM ALL, YOU ARE A GOD!!

Imagine a 15kg block moving with a speed of 20m/s. Calculate the kinetic energy of this block.

SSSSS [86.1K]

The kinetic energy formula is;

$KE=\frac{m.v^2}{2}$

The variable $m$ represents the mass. Its unit is kilogram. We are informed that the mass of the object is $15kg$ . The variable $v$ represents the linear velocity. Its unit is meter per second. We are informed that the linear velocity of the object is $20m/s$ . Let's find the kinetic energy of the object by substituting these values in the formula.

$KE=\frac{15kg.(20m/s)^2}{2}$
$2.KE=15kg.400m^2/s^2$
$2.KE=(6000)kgm^2/s^2$
$KE=(3000)kgm^2/s^2=3000J$

5 0

7 months ago

Which of the following molecules is diamagnetic according to molecular orbital theory? A) N₂⁺ B) N₂⁻ C) CN D) CN⁻

Gnom [1K]

Answer:

D) CN⁻

Explanation:

Hund's Rule of Maximum Multiplicity state that electrons go into degenerate orbitals of sub-levels (p,d, and f ) singly before pairing commences. Hund's rule is useful in determining the number of unpaired electrons in an atom. As such, it explains some magnetic properties of elements.

An element whose atoms or molecules contain unpaired electrons is paramagnetic. i.e., weakly attracted to substances in a magnetic field.

On the other hand, the element whose atoms or molecules are filled up with paired electrons is known as diamagnetic, i.e., not attracted by magnetic substances.

According to the molecular orbital theory, the diamagnetic molecule is CN⁻ because of the absence of unpaired electrons.

6 0

2 years ago

According to the website of the National Aeronautics and Space Administration (NASA), the average temperature of the universe is

charle [14.2K]

Answer : The temperature in degree Celsius is, $-270.45^oC$

Explanation :

The conversion used for the temperature from Kelvin to degree Celsius is:

$^oC=K-273.15$

where,

$K$ = temperature in Kelvin

$^oC$ = temperature in centigrade

As we are given the temperature in Kelvin is, 2.7

Now we have to determine the temperature in Kelvin.

$^oC=K-273.15$

$^oC=2.7-273.15$

$^oC=-270.45$

Therefore, the temperature in degree Celsius is, $-270.45^oC$

3 0

2 years ago

Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50

devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C $_{V}$ You can work out C $_{V}$

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC $_{V}$ ∆T

Polyatomic gas: C $_{V}$ = 3R

∆U= nC $_{V}$ ∆T

∆U= 28g x C $_{V}$ x (350K - 58.3K)

∆U = 28C $_{V}$ x 291.7

∆U = 10967.6 x C $_{V}$

5 0

2 years ago

Look at the picture below. Name the isotope

andrew11 [14]

It is potassium 39 definately

6 0

2 years ago