Answer:
PULLIN OUT THE COUPE AT THE LOT?
Explanation:
TOLD EM F*&*& 12 F*(&*(& SWAT
Answer:
The correct answer is - 4.
Explanation:
As we known and also given that the total of the superscripts that is mass numbers, A in the reactants and products must be the same.The mass of products A can understand and calculated by this -
The sum of the product mass number of products = mass of reactant
237Np93 →233 Pa91 +AZX is the equation,
Solution:
Mass of reactants = 237
Mass of products are - Pa =233 and A = ?
233 + A = 237
A = 237 - 233
A = 4
So the equation will be:
237Np93 →233 Pa91 +4He2 (atomic number Z = 2 ∵ difference in the atomic number of reactant and products)
Answer:
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Explanation:
tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=
mass compound
mass H
×100%
\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=
mass compound
mass C
×100%
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=
10.0g compound
2.5g H
×100%=25%
\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=
10.0g compound
7.5g C
×100%=75%
Arrhenius equation is LaTeX: y=-15035x+41.45y = − 15035 x + 41.45. What is the activation energy for this reaction is described below.
Explanation:
When the rates are equal and the concentration of reactants and products are constant not equal. Other like activation energies and potential energies are not necessary for equilibrium.
The activation energies of the forward and reverse reactions ,the rates of the forward and reverse reactions are 25 minutes ago.
For a hypothetical reaction the linear Arrhenius equation is LaTeX: y=-15035x+41.45y = − 15035 x + 41.45
15kj/mol is the answer..
