Write the formulas of the following compounds:
(a) barium chloride
(b) magnesium-nitride
(c) sulfur dioxide
(d) nitrogen trichloride
(e) dinitrogen trioxide
(f) tin(iv) chloride

(a) barium chloride :

(b) magnesium-nitride:

(c) sulfur dioxide :

(d) nitrogen trichloride :

(e) dinitrogen trioxide :

(f) tin(IV) chloride :

Explanation:

The nomenclature of ionic compounds is given by:

1. Positive is written first followed by the oxidation state of metal in roman numerals in square brackets.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

The naming of covalent compound is given by:

1. The less electronegative element is written first.

2. The more electronegative element is written second. Then a suffix is added with it. The suffix added is '-ide'.

If atoms of an element is greater than 1, then prefixes are added which are 'mono' for 1 atom, 'di' for 2 atoms, 'tri' for 3 atoms and so on.

a) Barium chloride is an ionic compound and the chemical formula will be

b)  Magnesium-nitride is an ionic compound and the chemical formula will be

c) Sulfur dioxide  is a covalent compound and the chemical formula will be

d) Nitrogen trichloride is a covalent compound and the chemical formula will be

e) Dinitrogen trioxide is a covalent compound and the chemical formula will be

f)  Tin(IV) chloride is an ionic compound and the chemical formula will be

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