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crimeas [40]
3 years ago
10

How does atomic radius affect coulombic attraction?

Chemistry
1 answer:
fomenos3 years ago
7 0
According to Coulomb's Law, as the atomic number increases within a series of atoms, the nuclear attraction for electrons will also increase, thus pulling the electron(s) closer to the nucleus.
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Acetone can be easily converted to isopropyl alcohol by addition of hydrogen to the carbon–oxygen double bond. calculate the ent
Vanyuwa [196]
<span>Answer: Enthalpy is delta-H- We need to look at the molecule and determine which bonds are broken adn which bonds are formed. Bonds that are broken are H-H (from the H2 molecule) and the C=O from acetone. their energies add up like this: 436 kJ + 745 kJ = 1181 kJ looking at the bonds formed, these are C-O, O-H, and C-H. these add up to 1229 kJ solving for delta H by taking the sum of the broken bonds and subtracting the sum of the formed bonds, like so: 1181 - 1229 = -48 kJ</span>
4 0
4 years ago
HELP<br> FREE BRAINLIEST
liberstina [14]
The universe comes into existence is first
The first neutral atoms form is second
The universe begins expanding is third
Gases form that will later go to shape stars and galaxies is fourth
Atomic nuclei form is last

I'm almost certain that is correct. Do not take my word for this.
6 0
3 years ago
Read 2 more answers
What is the mass of 1 atom?
VLD [36.1K]

6.7 mass because 1 atom equals 6.3 but if u add 4 it would be 6.7

6 0
3 years ago
If the mass of the object below is 28g, what is the density of the object below. Units are in cm below. Please round your answer
Murljashka [212]

Answer:

d = 0.93 g/cm³

Explanation:

Given data:

Mass of object = 28 g

Volume of object = 3cm×2cm×5cm

density of object = ?

Solution:

Volume of object = 3cm × 2cm ×5cm

Volume of object = 30 cm³

Density of object:

d = m/v

by putting values,

d = 28 g/ 30 cm³

d = 0.93 g/cm³

3 0
3 years ago
Water is being pumped from the bottom of a well 150 feet deep at a rate of 200 gal/hour into a vented storage tank 30 feet above
KengaRu [80]

Explanation:

As the given data is as follows.

        Height, H = 150 feet

 Heat gain = 30,000 BTU/hr,  and  Heat loss = 25000 BTU/hr

  m = mass of water heated = 700 gallons = 5810 lbs

C_{p} is the heat capacity of water = 1 BTU/lb ^{o}F (given)

      \Delta T = temperature difference = 120^{o}F - 35^{o}F

Heat energy required to heat 700 gal can be calculated as follows:

    Heat Required = 5810 lbs \times 1 BTU/lb^{o}F \times (120^{o}F - 35^{o}F)

Thus, water rises till 120^{o}F.

3 0
4 years ago
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