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Allushta [10]
2 years ago
12

Lunar Eclipse. Choose ALL that apply

Chemistry
2 answers:
antoniya [11.8K]2 years ago
6 0

Answer:

b and a i jus did it

Explanation: none

Ronch [10]2 years ago
5 0

Answer:

the answer to 1 is all the answers

the 2 one is Moon moves between the sun and Earth and The moon blocks out the light of the sun and casts a shadow onto the Earth  

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Diamond has a density of 3.513 g/cm³. The mass of diamonds is often measured in "carats," where 1 carat equals 0.200
Gennadij [26K]

The volume of a 1.86-carat diamond in cubic centimeters is 0.106 cm³

Given,

The density of a diamond is  3.513 g/cm³.

We have to find out the volume of a 1.86-carat diamond in cubic centimeters.

Convert the units of the diamond from carat to grams, we have:

(1.86 carats) x (0.200 g / 1 carat) = 0.372 g

The volume of the diamond is obtained by dividing the mass by the density, therefore using the formula, we get

                            v = m / d

                            v = 0.372 g / (3.51 g/cm³) = 0.1059 cm³

                       or, v = 0.106 cm³ (approx)

Therefore, the volume of a 1.86-carat diamond is approximately 0.106 cm³.

To learn more about the volume, visit: brainly.com/question/1578538

#SPJ9

4 0
2 years ago
Describe the electrons in an atom of carbon in the ground state. Your response must
zepelin [54]

Based on the information given, it should be noted that the ground-state electron configuration of carbon is 1s2 2s2 2p2.

<h3>What is an electron?</h3>

Electrons are simply the subatomic particles which orbit the nucleus of an atom.

The arrangement of electrons in the atomic orbitals of an atom is known as the electron configuration. This can be determined by using a periodic table.

It should be noted that carbon is the sixth element with a total of 6 electrons in the periodic table. Thus, the atomic number Z = 6.

In conclusion, the ground-state electron configuration of carbon is 1s2 2s2 2p2.

Learn more about carbon on:

brainly.com/question/105003

8 0
2 years ago
The sun radiates different type of Electromagnetic energy (TRUE or FALSE)
Paha777 [63]
That statment is true
7 0
3 years ago
g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The heig
kumpel [21]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, \Delta S_{env} = -1.18 J/K

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

Learn more: brainly.com/question/22655760

6 0
2 years ago
What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C
Digiron [165]

Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1= initial pressure at 26.5^oC = ?

P_2 = final pressure at -5.1^oC = 100 mm Hg

= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol

R = gas constant = 8.314 J/mole.K

T_1= initial temperature = 26.5^oC=273+26.5=299.5K

T_2 = final temperature =-5.1^oC=273+(-5.1)=267.9K

Now put all the given values in this formula, we get

\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]

\log  (\frac{P_1}{100})=-0.576

\frac{P_1}{100}=0.265

P_1=26.5mmHg

Thus the vapor pressure of CS_2CS_2 in mmHg at 26.5 ∘C is 26.5

7 0
3 years ago
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