**Answer: The standard enthalpy of formation of is -252.1 kJ/mol.**

**Explanation:**

The balanced chemical reaction is,

The expression for enthalpy change is,

Putting the values we get :

**Thus standard enthalpy of formation of is -252.1 kJ/mol.**

**Answer:**

Mg²⁵ = 10.00%

Mg²⁶ = 45.04%

Mg²⁴ = 44.96%

**Explanation:**

**Given data:**

Atomic mass of Mg²⁶ = 25.983

Atomic mass of Mg²⁵ = 24.986

Atomic mass of Mg²⁴ = 23.985

Abundance of Mg²⁵ = 10.00%

Abundance of Mg²⁶ = ?

Abundance of Mg²⁴ = ?

**Solution:**

Average atomic weight of Mg =** **25.983 + 24.986+ 23.985 / 3

Average atomic weight of Mg = 74.954/3

Average atomic weight of Mg = 24.985 amu

**Abundance of **

Mg²⁵ = 10.00

Mg²⁶ = x

Mg²⁴ = 100- 10 - x = 90 -x

**Formula:**

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100

24.985 = (0.1×24.986)+(90-x×23.985) + ( x ×25.983 ) /100

24.985 = 249.86 + 2158.65 - 23.985x + 25.983x / 100

24.985 = 2408.51 + 1.998 x / 100

2498.5 = 2408.51 + 1.998 x

1.998 x = 2498.5 - 2408.51

1.998 x = 89.99

x = 89.99 /1.998

x = 45.04

Now we put the value of x:

Mg²⁵ = 10.00

Mg²⁶ = x (45.04)

Mg²⁴ = 90 -x (90 - 45.04 = 44.96)

**Answer:**

I don't know if this will help but. The Great Sphinx has suffered from erosion and vandalism. Hatshepsut's stepson was too young to rule at the time of her husband's death, so she just took the job.

**Answer:**

i would happily help but i don't understand that language sorry

**Answer:**

ΔS > 0 only for choice E: CH4(g) + H2O (g) → CO(g) + 3 H2(g)

**Explanation:**

Our strategy in this question is to use the trend in entropies :

S (solids) less than S (liquids) less than S (gases)

Also we have to look for the molar quanties involved of each state and their change to answer the question:

A. N2(g) + 3 H2(g) → 2 NH3(g)

Here we have 4 moles gases going to 2 moles of products, so the change in entropy is negative.

B. Na2CO3(s) + H2O(g) + CO2(g) → 2 NaHCO3(s)

The change in entropy is negative since we have 2 mol gases in the reactants and zero in the products.

C. CH3OH(l) → CH3OH(s)

A liquid has a higher entropy than a solid so ΔS is negative

D. False see A,B,C

E. The change in moles of gases is 4 - 2= 2, therefore ΔS is greater than O.