full octets live in noble gas world so you need to be whatever your charge is away from there in steps on the periodic table so s-1 is the one that falls short as it is 2 moves away from Ar which is its closest Noble gas

Explanation:

6 0

3 years ago

What is the concentration in mol L-l of a 12% solution of tetrahydrate magnesium chloride (MgCl2)?

madreJ [45]

The concentration in mol L-l (M) = 0.717

<h3>Further explanation</h3>

Given

12% solution of tetrahydrate magnesium chloride (MgCl₂)

Required

The concentration

Solution

Tetrahydrate magnesium chloride (MgCl₂)⇒MgCl₂.4H₂O

MW = Ar Mg+2. Ar Cl+8. Ar H + 4. Ar O

MW=24.31 + 2 x 35.45 + 8 x 1.01 + 4 x 16

MW=24.31+70.9+8.08+64

MW=167.29 g/mol

12% solution = 12 % m/v = 12 g in 100 ml solution

For 1 L solution :

$\tt \dfrac{1}{0.1}\times 12~g=120~g$

The concentration in g/L = 120 g/L

Convert grams to moles :

$\tt mol=\dfrac{120}{167.29}=0.717$

3 0

3 years ago

When a 3.80 g sample of C8H18(l) is burned in a bomb calorimeter, the temperature of the calorimeter rises by 27.3 oC. The heat

Fudgin [204]

<u>Answer:</u> The enthalpy of the reaction is -5112.5 kJ/mol

<u>Explanation:</u>

To calculate the heat absorbed by the calorimeter, we use the equation:

$q=c\Delta T$

where,

q = heat absorbed

c = heat capacity of calorimeter = 6.18 kJ/°C

$\Delta T$ = change in temperature = 27.3°C

Putting values in above equation, we get:

$q=6.18kJ/^oC\times 27.3^oC=168.714kJ$

Heat absorbed by the calorimeter will be equal to the heat released by the reaction.

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of octane = 3.80 g

Molar mass of octane = 114 g/mol

Putting values in above equation, we get:

$\text{Moles of octane}=\frac{3.80g}{114g/mol}=0.033mol$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta E=\frac{q}{n}$

where,

q = amount of heat released = -168.714 kJ

n = number of moles = 0.033 moles

$\Delta E$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta E=\frac{-168.714kJ}{0.033mol}=-5112.5kJ/mol$

Hence, the enthalpy of the reaction is -5112.5 kJ/mol

4 0

3 years ago