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sergejj [24]
4 years ago
10

I need help with Kepler’s law!

Chemistry
1 answer:
strojnjashka [21]4 years ago
7 0
In astronomy, Kepler's laws of planet motion are three scientific laws describing the motion of planets around the Sun, published by Johannes Kepler between 1609 and 1619.
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Determine the mass of 2.5 cups of water if the density of water is 1.00 g/cm3 and 1 cup = 240 ml. i
Radda [10]

The mass is simply the product of volume and density. But first, let us convert the volume into cm^3 (cm^3 = mL):

volume = 2.5 cups * (240 mL/cup)

volume = 600 mL = 600cm^3

 

So the mass is:

mass = 600 cm^3 * (1 g / cm^3)

<span>mass = 600g</span>

6 0
4 years ago
What is the mass of an atom with seven protons, seven neutrons, and seven electrons? Record and bubble in your answer below.
Irina-Kira [14]

Answer:

Mass would be 14 amu

Explanation:

7 0
3 years ago
Read 2 more answers
Which has the strongest intermolecular bonds? liquid gas solid water
Delicious77 [7]
It would be solid
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3 0
4 years ago
Determine the number of unpaired electrons expected for [Fe(NO2)6]3−and for [FeF6]3− in terms of crystal field theory.
arsen [322]

Answer:

A. One unpaired electron

B. 5 unpaired electrons

Explanation:

In A ,Fe is in +3 oxidation state and Electronic configuration- [Ar]3d5

And NO2 is a strong field ligand hence it causes pairing in t2g orbitals and results one unpaired electron in dZX orbital.

In B, also Fe is in +3 oxidation state but F is weak field ligand hence causes no pairing of Electrons hence it results 5 unpaired electrons with electronic configuration t2g^3 eg^2

7 0
3 years ago
Ask Your Teacher A 1.04-mole sample of ammonia at 11.0 atm and 25°C in a cylinder fitted with a movable piston expands against a
RUDIKE [14]

Answer:

Explanation:

a) Volume of the gas nRT / P

= 1,04 X 8.3 X 298 / 11 X 10⁵ m³

= 233.85 x 10⁻⁵ m³

= 233 x 10⁻² L

2.33 L

P₁V₁ / T₁ =P₂V₂/T₂

(11 X 2.33) / 298 = (1 X 24.2) / T

T = 281.37 K

= 8.37 degree

b ) w = p x change in volume

= 10⁵ x ( 24.2 - 2.33 ) x 10⁻³ J

= 21.87 X 10² J

2187 J

q  = n x Cp x (25 - 8.37 )

= 1.04 x 35.66x 16.63 J

= 616.65 J

ΔU = Q - W

= 616.65  - 2187 J

= - 1570.35 J

=

=

7 0
3 years ago
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