I think it’s CTell me if I’m wrong
Answer:
The new volume of the gas is 276.45 mL.
Explanation:
Charles's law indicates that for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases, and as the temperature decreases, the volume of the gas decreases.
Charles's law is a law that mathematically says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

Analyzing an initial state 1 and a final state 2, it is satisfied:

In this case:
- V1= 250 mL
- T1= 293 K
- V2= ?
- T2= 324 K
Replacing:

Solving:

V2= 276.45 mL
<em><u>The new volume of the gas is 276.45 mL.</u></em>
The elements in<span> groups 1a through 7a are called d. representative elements. The periodic table is organized according to increasing order of the atomic number of the elements. Groups have similar properties while the period of the elements determines the valence electrons in the outer shells.</span>
Answer: a)
: Decomposition
b)
: double displacement
c)
: Synthesis (Combination)
d)
: redox
Explanation:
Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

A double displacement reaction is one in which exchange of ions take place.

Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.

Redox reaction is a type of chemical reaction in which oxidation and reduction takes place in one single reaction. The oxidation number of one element increases and the oxidation number of other element decreases.

Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
______
NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.