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Hatshy [7]
3 years ago
12

the gravity Neptune is about 1.1 times the gravity of earth.how will the mass of an object on Neptune compare with its mass on e

arth
Chemistry
1 answer:
Darina [25.2K]3 years ago
5 0

Answer:

It will be exactly the same.

Explanation:

Mass is the amount of matter in an object. Mass is constant.

Weight is the gravitational force of attraction for an object. The weight of an object on Neptune is 1.1 times its weight on Earth

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Humans opinions of evolution.
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3 years ago
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How many calories are absorbed in a process that absorbs 0.128 joules?
earnstyle [38]

Answer:

There are 0.0305 calories in 0.128 joules

Explanation:

Given that,

Heat absorbed, Q = 0.128 J

We need to find the heat energy absorbed in calories.

We know that the relation between joules and calories is as follows :

1 calorie = 4.184 J

1 J = (1/4.184) J

So,

0.128\ J=\dfrac{0.128 }{4.184}\\\\=0.0305\ cal

So, there are 0.0305 calories in 0.128 joules

7 0
3 years ago
In the number 48.93, which digit is estimated?
Fudgin [204]
The answer is 3.

Explanation:
It’s the last number and it can’t be 9 because then it would be 48.9 and no 3.
7 0
3 years ago
P-fluoroanisole reacts with sulfur trioxide and sulfuric acid. Draw the major product of this substitution reaction; if applicab
Ipatiy [6.2K]

Answer: (Structure attached).

Explanation:

This type of reaction is an aromatic electrophilic substitution. The overall reaction is the replacement of a proton (H +) with an electrophile (E +) in the aromatic ring.

The aromatic ring in p-fluoroanisole has two sustituents, an <u>halogen</u> and a <u>methoxy group</u>, which are <em>ortho-para</em> directing substituents.

Aryl sulfonic acids are easily synthesized by an electrophilic substitution reaction aromatic using <u>sulfur trioxide as an electrophile</u> (very reactive).

The reaction occurs in three steps:

  1. The attack on the electrophile forms the sigma complex.
  2. The loss of a proton regenerates an aromatic ring.
  3. The sulfonate group can be protonated in the presence of a strong acid (H₂SO₄).

Normally, a mixture of <em>ortho-para</em> substituted products would be obtained. However, since both <em>para</em> positions are occupied, only the <em>ortho </em>substituted product is obtained here.

8 0
3 years ago
A certain liquid sample has a volume of 14.7 mL and a mass of 22.8 grams. Calculate the density.
ira [324]
M = 22.8 g
V = 14.7 mL
ρ - ?

ρ = m/V
ρ = 22.8/14.7 = 1.55 g/mL

4 0
3 years ago
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