Answer:
Different materials will change temperature at different rates when exposed to the same amount of thermal energy. This is because each substance has its own specific heat.
Explanation:
that is the answer
Answer:
As
Explanation:
For any element to exhibit the pattern of ionization energy shown in the question, it must possess five electrons in its outermost shell. These five electrons are not lost at once. They are lost progressively until the valence shell becomes empty. The ionization energy increases steadily as more electrons are lost from the valence shell.
The only pentavalent element among the options in arsenic, hence the answer.
The trick for this problem is to understand atomic mass: the fact that different atoms have different masses. What we need to do is add up all the atomic masses of the compound and work out the ratio of mass of water to the mass of sodium carbonate. Atomic masses are often given for each atom in the periodic table, but you can look them up on google too.
You can do this by adding up individual atoms for each molecule, or you can shortcut and lookup the molar mass of the compound (i.e.the task already done for you).
The molar mass of water is 18.01g/mole so for 10 moles of water we have a mass of 180.1g.
The molar mass of sodium carbonate is 106g/mole (google).
So the total mass of the sodium carbonate decahydrate compound is 180.1+106 = 286.1g, of which water would make up 180.1g, so the percentage of water is is 180.1/286.1 = 0.629, so we can round this to 63%
:)
Answer:
b) 0.47
Explanation:
MwC5H12 = 72.15g/mol
⇒mol C5H12 = (10.0)*(mol/72.15)=0.1386molC5H12
MwC6H14=86.18g/mol
⇒molC6H14=(20.0)*(mol/86.18)=0,232
MwC6H6=78.11g/mol
⇒molC6H6=(10.0)*(mol/78.11)=0.128molC6H6
<h3>XC6H14=(0.232)/(0.1386+0.232+0,128)=0.465≅0.47</h3>
Answer:
Maintaining a high starting-material concentration can render this reaction favorable.
Explanation:
A reaction is <em>favorable</em> when <em>ΔG < 0</em> (<em>exergonic</em>). ΔG depends on the temperature and on the reaction of reactants and products as established in the following expression:
ΔG = ΔG° + R.T.lnQ
where,
ΔG° is the standard Gibbs free energy
R is the ideal gas constant
T is the absolute temperature
Q is the reaction quotient
To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.
