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mezya [45]
4 years ago
8

What is the moden periodic law​

Chemistry
1 answer:
Zarrin [17]4 years ago
5 0

Answer:

Here's what I get  

Explanation:

The modern periodic law states that the physical and chemical properties of the elements are periodic functions of their atomic numbers.

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The area surrounding the nucleus is called: protons electrons neutrons electron cloud
vesna_86 [32]

Answer:

eletrons

Explanation:

eletrons is not in the neuclus its around it

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3 years ago
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If a sample of oxygen occupies a volume of 2.15 L at a pressure of 58.0 kPa and a temperature of 25°C, what volume would this sa
Anna35 [415]

The answer is A) 1.13 L

8 0
3 years ago
What is a materials ability to dissolve in a solvent
ExtremeBDS [4]
I believe it might be A, hope this helps!
5 0
3 years ago
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B. What useful functions do oxidation numbers
disa [49]

Explanation:

b. What useful functions do oxidation numbers  serve?

It is used to show oxidation and reduction (loss and gain of electrons)

b. How many molecules are in 1 mole of  molecules?

1 mole = 6.022 * 10^23 molecules

c. What is the name given to the number of  molecules in 1 mole?

Avogadro's Number of molecules

21. a. What is the molar mass of an element?

This is the mass of an element divided by the number of moles.

Molar mass = Mass / Number of moles

b. Write the molar mass rounded to two  decimal places of carbon, neon, iron and  uranium.

amu = Atomic Mass Unit

Carbon = 12.01 amu

Neon = 20.18 amu

Iron = 55.85 amu

Uranium = 238.03 amu

7 0
3 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
3 years ago
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