Chemistry, 17.12.2020 18:30, 20170020

Which is a component of John Dalton’s atomic theory? A. The ratio of atoms in a compound is fixed.
B. The atoms of different elements are the same.
C. An atom is a small particle of matter that can be broken down.
D. A reaction can create or destroy atoms as well as rearrange them
(It's either A or C but I am not sure)

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All matter is made of atoms. Atoms are indivisible and indestructible. 3) Compounds are formed by a combination of two or more different kinds of atoms. 4) A chemical reaction is a rearrangement of atoms.

if that didnt help, this will. The English scientist John Dalton were one of the early scientists who paved the way to science, particularly in atomic chemistry. His atomic theory jumpstarted further developments made by J.J Thomson, Rutherford, Bohr, Schrodinger and many more.

Dalton's atomic theory can be summarized in 5 key points:

1. Atoms are tiny, indivisible and indestructible particles of an element.

2. Atoms of different elements vary in size and mass.

3. Atoms of an element are uniform in size and mass.

4. Compounds are formulated by a fixed ratio of individual atoms of elements.

5. Reactions undergo by rearranging the atoms of the reactants to form the products.

From the choices, the acceptable answer would be: The ratio of atoms in a compound is fixed.

Explanation:

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answered: Guest

answer: enthalpy

explanation:

for a exothermic reaction, the energy is released i.e. the energy of the products is lower than the energy of the reactants.

$\delta h=e_p-e_r$

$\delta h$ = enthaply

$e_p$ = energy of products

$e_r$ = energy of reactants

as $e_p< e_r$ , thus $\delta h$ comes out to be negative. negative value of enthalpy shows that the reaction is exothermic.

answered: Guest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.