Chemistry, 13.10.2019 01:50 aaqa3735

Which of these elements has four valence electrons? hafnium radon silicon sulfur

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Answer from: jahzz

Answer : The elements that has four valence electrons are hafnium (Hf) and silicon (Si).

Explanation :

Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom are determined by the electronic configuration.

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

Valence electrons or outermost electrons : They are electrons in an atom that participate in bonding.

(a) Hafnium is a metal that belongs to group 4 and period 6. The atomic number of Hafnium is, 72.

The electronic configuration of hafnium will be,

$[Xe]4f^{14}5d^26s^2$

The total number of valance electrons = [2 + 2] = 4

(b) Radon is a noble gas that belongs to group 18 and period 6. The atomic number of radon is, 86.

The electronic configuration of radon will be,

$[Xe]4f^{14}5d^{10}6s^26p^6$

The total number of valance electrons = [2 + 6] = 8

The electronic configuration of silicon will be,

1s^22s^22p^63s^23p^2

The total number of valance electrons = [2 + 2] = 4

(d) Sulfur is a non-metal that belongs to group 16 and period 3. The atomic number of sulfur is, 16.

The electronic configuration of sulfur will be,

1s^22s^22p^63s^23p^4

The total number of valance electrons = [2 + 4] = 6

Hence, from this we conclude that, the elements that has four valence electrons are hafnium (Hf) and silicon (Si).

Answer from: Quest

is there multiple choice answers?

Answer from: Quest

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

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