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3 years ago

Which metal in Period 5 is very reactive and has two valence electrons in each atom?

2 answers:

8 0

The answer is Strontium(Sr). The reactive increase from right to left. And this element has two valence electrons. So Rb is not correct. Then the very reactive metal is Sr.

irakobra [83]3 years ago

8 0

<u>Answer:</u> The correct answer is Strontium.

<u>Explanation:</u>

Reactivity of an element is defined as the tendency to loose or gain electrons. Reactivity decreases as we move from left to right in a period and it increases as we move from up to down in a group.

A metal which belongs to Period 5 and has 2 valance electrons is Strontium. This element belongs to Group 2.

The electronic configuration of strontium (Sr) = $[Kr]5s^2$

Hence, the correct answer is Strontium.

You might be interested in

How many moles of C2H2 are needed to react completely with 84 mol O2

Butoxors [25]

Balanced chemical equation:

2 C2H2 + 5 O2 = 4 CO2 + 2 H2O

2 moles C2H2 ---------------- 5 moles O2
moles C2H2 ------------------ 84 moles O2

moles C2H2 = 84 * 2 / 5

molesC2H2 = 168 / 5 => 33.6 moles of C2H2

4 0

3 years ago

_Fe2O3 + 2CO —> _Fe + _CO2

soldi70 [24.7K]

<h3>Answer:</h3>

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

<h3>Explanation:</h3>

Concept tested: Balancing of chemical equations

A chemical equation is balanced by putting appropriate coefficients on the products and reactants of the equation.
Balancing chemical equations ensures that chemical equations obey law of conservation of mass.
In this case; to balance the above equation we put the coefficients, 1, 3, 2, and 3 on the reactants and products.
Therefore; the balanced chemical equation for the reaction is;

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

8 0

3 years ago

At a particular temperature, 12.0 moles of so3 is placed into a 3.0-l rigid container, and the so3 dissociates by the reaction 2

saul85 [17]

2 SO₃ --> 2 SO₂ + O₂
I 12 0 0
C -2x +2x +x
---------------------------------------------
E 12-2x 2x x

Since the moles of SO₂ at equilibrium is 3 mol, 2x = 3. Then, x = 1.5 mol. So, the amounts at equilibrium is:
SO₃: 12 - 2(1.5) = 9
SO₂: 2(1.5) = 3
O₂: 1.5

The formula for K basing on the stoichiometric reaction is:
K = [SO₂]²[O₂]/[SO₃]²
where the unit used is conc in mol/L.

K = [3 mol/3 L]²[1.5 mol/3 L]/[9 mol/3 L]²
<em>K = 0.0556</em>

4 0

3 years ago

Consider an amphoteric hydroxide, m(oh)2(s), where m is a generic metal. estimate the solubility of m(oh)2 in a solution buffere

Colt1911 [192]

Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7
∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
= (2x10^-16)/(1x10^-7)^2
= 0.02 M
b) at PH =10
when POH = 14- PH = 14-10 = 4
when POH = -㏒[OH-]
    4 = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
= 2x10^-16 / (1x10^-4)^2
  = 2x10^-8 M
c) at PH= 14
when POH = 14-PH
= 14 - 14
= 0
when POH = -㏒[OH]
  0 = - ㏒[OH]
∴[OH] = 1 m
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
= (2x10^-16) / 1^2
= 2x10^-16 M

8 0

3 years ago

Joseph drives 25miles North in 15 minutes then he drives 40 miles east in 30 minutes, then he uses cruise control option for 1 h

Musya8 [376]

Answer: 1.25 miles per minute

Explanation:

Average speed is the rate of change of total distance covered per unit time.

i.e Average speed = (Total distance / Time taken)

Total distance covered = (25miles + 40 miles + 70 miles + 15 miles)

= 150 miles

Total time taken = ( 15 minutes + 30 minutes + 1 hour + 15 minutes) = 120 minutes

Since 60 minutes = 1 hour, the total time taken is 120 minutes

Now, apply Average speed = (Total distance / Time taken)

= (150 miles / 120 minutes)

= 1.25 miles per minutes

Thus, Joseph drove with an average speed of 1.25 miles per minute.

7 0

3 years ago