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3 years ago

Electron configuration for be+2

Chemistry

1 answer:

svlad2 [7]3 years ago

6 0

Answer:

atomic number of Be =4

electronic configuration :1s²2s²

but for Be2+ atomic number = 4-2 =2

electronic configuration : 1s²

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How is the use of word theory in science different from its more common use

weqwewe [10]

It is like an idea that they come up with in an experimeant

4 0

3 years ago

PCl3(g) + Cl2(g) ⇋ PCl5(g) Kc = 91.0 at 400 K. What is the [Cl2] at equilibrium if the initial concentrations were 0.24 M for PC

Dmitry_Shevchenko [17]

Answer:

[Cl₂] in equilibrium is 1.26 M

Explanation:

This is the equilibrium:

PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)

Kc = 91

So let's analyse, all the process:

PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)

Initially 0.24 M 1.50M 0.12 M

React x x x

Some amount of compound has reacted during the process.

In equilibrium we have

0.24 - x 1.50 - x 0.12 + x

As initially we have moles of product, in equilibrium we have to sum them.

Let's make the expression for Kc

Kc = [PCl₅] / [Cl₂] . [PCl₃]

91 = (0.12 + x) / (0.24 - x) ( 1.50 - x)

91 = (0.12 + x) / (0.36 - 0.24x - 1.5x + x²)

91 (0.36 - 0.24x - 1.5x + x²) = (0.12 + x)

32.76 - 158.34x + 91x² = 0.12 +x

32.64 - 159.34x + 91x² = 0

This a quadratic function:

a = 91; b= -159.34; c = 32.64

(-b +- √(b² - 4ac)) / 2a

Solution 1 = 1.5

Solution 2 = 0.23 (This is our value)

So [Cl₂] in equilibrium is 1.50 - 0.23 = 1.26 M

5 0

3 years ago

Will give 20 points and brainliest help

Blababa [14]

the answer you have chosen is correct

8 0

3 years ago

Consider the reaction directly below and answer parts a and b. C6H4(OH)2 (l) + H2O2 (l) à C6H4O2 (l) + 2 H2O (l)

dedylja [7]

Answer:

a. -206,4kJ

b. Surroundings will gain heat.

c. -115kJ are given off.

Explanation:

It is possible to obtain ΔH of a reaction using Hess's law that consist in sum the different ΔH's of other reactions until obtain the reaction you need.

Using:

(1) C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ

(2) H₂(g) + O₂(g) → H₂O₂ (l) ΔH: -187.8 kJ

(3) H₂(g) + 1/2O₂(g) → H₂O(l) ΔH: -285.8 kJ

It is possible to obtain:

C₆H₄(OH)₂(l) + H₂O₂(l) → C₆H₄O₂(l) + 2H₂O(l)

From (1)-(2)+2×(3). That is:

(1) C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ

-(2) H₂O₂(l) → H₂(g) + O₂(g) ΔH: +187.8 kJ

2x(3) 2H₂(g) + O₂(g) → 2H₂O(l) ΔH: 2×-285.8 kJ

The ΔH you obtain is:

+177,4kJ + 187,8kJ - 2×285.8 kJ = -206,4kJ

b. When ΔH of a reaction is <0, the reaction is exothermic, that means that the reaction produce heat and the surroundings will gain this heat.

c. 20,0g of H₂O are:

20,0g× $\frac{1mol}{18,01g}$ = 1,11 mol H₂O

As 2 moles of H₂O are produced when -206,4kJ are given off, when 1,11mol of H₂O are produced, there are given off:

1,11mol H₂O× $\frac{-206,4kJ}{2mol}$ = -115kJ

I hope it helps!

8 0

3 years ago

Does sodium (Na) and argon (Ar) on the periodic table have similar properties? Explain in terms of location on the periodic tabl

iVinArrow [24]

Sodium (Na) and Argon (Ar) on the periodic table does not have similar properties.

PERIODIC TABLE:

Periodic table is the organized table that showcases the position of elements. The periodic table is organized into groups and periods.

Groups of the periodic table is in a vertical format labeled 1 - 7 while the period is in a horizontal format labeled 1 - 7.

Elements in the same group have the same number of valence electrons and hence, possess similar chemical properties.

Sodium (Na) belongs to group 1 while argon is a noble gas, hence, they are not in the same group of the periodic table and cannot have same properties.

Learn more at: brainly.com/question/24089206?referrer=searchResults

6 0

3 years ago