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3 years ago

What is the collection of protons and neutrons in the center of an atom called

Chemistry

2 answers:

sdas [7]3 years ago

8 0

If I am correct it should be nucleons

Irina18 [472]3 years ago

8 0

The collection of protons and neutrons in the center of the atom is called the nucleus.

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Methylene chloride (CH2Cl2) has fewer chlorine atoms than chloroform (CHCl3). Nevertheless, methylene chloride has a larger mole

icang [17]

Answer:

Explanation has been given below.

Explanation:

Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.

3 0

4 years ago

Living things within an ecosystem

ohaa [14]

Living things included are organisms like animals, plants, bacteria and insects, as well as nonliving components like rocks, soil, sunlight and water.

Hope this helps!! would appreciate a brainliest!

7 0

3 years ago

The density of a liquid can be determined by weighing a known volume of the liquid in a graduated cylinder. the data from the ex

Kipish [7]

Answer:- density of the liquid is 0.21 gram per mL.

Solution:- $density=\frac{mass}{volume}$

From given data, mass of cylinder = 10.4 g

mass of cylinder along with liquid = 11.7 g

So, mass of liquid = 11.7 g - 10.4 g = 1.3 g

Volume of liquid = 6.2 mL

Let's divide the mass by the volume to get the density of the liquid:

$density=\frac{1.3g}{6.2mL}$

density = $0.21\frac{g}{mL}$

So, the density of the liquid is 0.21 gram per mL.

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3 years ago

Which book would a meteorologist most likely use as a reference?

TiliK225 [7]

Answer:

- Predicting the Weather

Hope this helps!

7 0

3 years ago

Read 2 more answers

A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate

inna [77]

Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams and dry-weight of 85.2 grams

a) wet density of the CL sample

wet density can be expressed as p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

p = 0.0453 lbs/in³

so the wet density of the CL sample is 0.0453 lb/in³

water content of sample is taken as;

w = (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 / 1.6537

Sd = 0.0274 lb/in³

therefore the dry density of the CL sample is 0.0274 lb/in³

8 0

3 years ago