The enthalpy change : -196.2 kJ/mol
<h3>Further explanation </h3>
The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
Reaction
2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)
∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂
Answer:
58.92 g EDTA
Explanation:
315.1 mL = .3151 L
M = Moles / Liter
.3151 L x <u>0.5 mol EDTA</u> x <u>374 g EDTA</u> = 58.92 g EDTA
1 L EDTA 1 mol EDTA
Answer:
3
Explanation:
Applying,
= R/R'............... Equation 1
Where n' = number of halflives that have passed, R = Original atom of the substance, R' = atom of the substance left after decay.
From the question,
Given: R = 40 atoms, R' = 5 atoms
Substitute these values into equation 1
= 40/5
= 8
= 2³
Equation the base,
n' = 3
The density of the block will be "7.11 gm/cm³". A further explanation is provided below.
Given values are:
Density of iron,
Mass of block of metal,
Volume of block,
By using the formula,
→ 
then,
→ The density of block will be:
= 
= 
= 
Learn more:
brainly.com/question/10804879
