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serious [3.7K]
3 years ago
9

The energy needed to remove an electron is called ionization energy.

Chemistry
1 answer:
salantis [7]3 years ago
7 0
<h2>                             Emma here<em>!</em></h2>

Answer:

im pretty sure its decreases going across a row (left to right)

Explanation:

As you move from left to right across a period on the periodic table the size of an atom will decrease

<h3>Hope this helps<em>! </em>Have a nice day<em>!</em></h3>
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What is a mutation?
Masja [62]

Answer:

the answer is A hope this helps

3 0
3 years ago
How many grams of chlorine gas can be produced when 50.0 grams of aluminum chloride decompose? 2AlCl3→ 2Al + 3Cl2
lianna [129]

Answer:

Explanation:

Approx.

425

⋅

g

Explanation:

2

A

l

(

s

)

+

3

C

l

2

(

g

)

→

2

A

l

C

l

3

(

s

)

You have given a stoichiometrically balanced equation, so bravo.

The equation explicitly tells us that

54

⋅

g

of aluminum metal reacts with

6

×

35.45

⋅

g

C

l

2

gas to give

266.7

⋅

g

of

aluminum trichloride

hope this helps

6 0
2 years ago
which count of subatomic particles gives an atom its unique identity as an element, and represents its atomic number? a. the num
Arada [10]
The subatomic particles that identifies an element and also represents its atomic number would be A. The number of protons.
7 0
2 years ago
Read 2 more answers
The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal
spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0
3 years ago
Astronomers studying the planet of Rhombus have detected sedimentary rock on its surface. One astronomer wonders if material in
Fed [463]

Answer:

igneous rock CAN become sedimentary rock through a process called ROCK CYCLE.

Explanation:

Rocks can be defined as solid structures of minerals that are formed naturally over a period of time. They are grouped into three main types which includes the following:

- igneous rock

- sedimentary rocks and

- metamorphic rocks.

Rocks are capable of transforming from one type to another through a process known as rock cycle. There are two forces that brings about this process which includes:

- The internal force : this is the Earth’s internal heat engine, which moves material around in the core and the mantle and leads to slow but significant changes within the crust.

- The external force: this is the the hydrological cycle, which is the movement of water, ice, and air at the surface, and is powered by the sun.

Molten magma cools to form either extrusive igneous rock or intrusive igneous rock. With time they undergo weathering, eroded, transported, and then deposited as sediments which are being compressed and cemented into SEDIMENTARY ROCKS. Again through the above mentioned forces, different kinds of rocks are either uplifted, to be re-eroded, or buried deeper within the crust where they are heated up, squeezed, and changed into METAMORPHIC ROCK.

Therefore the material in this sedimentary rock found in Rhombus planet used to be in igneous rock deep in Rhombus's interior due to continuous rock cycling on the planet. I hope this helps, thanks.

5 0
3 years ago
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