FOR 50 POINTS & A BRAINIST* 1.) For an atom of sulfur, there are
A.) two electron shells with 6 valence electrons
B.) three electron shells with 6 valence electrons
C.) four electron shells with 6 valence electrons
D.) five electron shells with 6 valence electrons

2.) Which atom has four electron shells with 7 valence electrons?
A.) F
B.) CI
C.) Br
D.) I

3.) Which atom has three electron shells with 8 electrons in its outermost orbit?
A.) He
B.) Ne
C.) Ar
D.) Kr

4.) Which method should be used to draw electrons that are involved in a collision during a chemical reaction?
A.) Bohr's model
B.) Lewis dot model
C.) Addition model
D.) Subtraction model

1. B

2. C

3. C

4. B

Explanation:

1. An atom of sulfur has an atomic number of 16. It has three electron shells and six valence electrons, therefore, it belongs to period 3, group VIA of the periodic table

2. Bromine has an atomic number of 35. It belongs to period 4 group VIIA of the periodic table. Therefore, it has four electron shells and seven valence electrons.

3. Argon has an atomic number of 18. It belongs to period 3, group IVA of the periodic table. Therefore, it has three electron shells and eight valence electrons.

4. The Lewis dot model use structures that represent the valence electrons of atoms within a molecule existing as lone pairs or when involved in bonding. They can be used to draw electrons that are involved in a collision during a chemical reaction.

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

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :