Answer:
Consider this balanced chemical equation:
2 H2 + O2 → 2 H2O
We interpret this as “two molecules of hydrogen react with one molecule of oxygen to make two molecules of water.” The chemical equation is balanced as long as the coefficients are in the ratio 2:1:2. For instance, this chemical equation is also balanced:
100 H2 + 50 O2 → 100 H2O
This equation is not conventional—because convention says that we use the lowest ratio of coefficients—but it is balanced. So is this chemical equation:
5,000 H2 + 2,500 O2 → 5,000 H2O
Again, this is not conventional, but it is still balanced. Suppose we use a much larger number:
12.044 × 1023 H2 + 6.022 × 1023 O2 → 12.044 × 1023 H2O
These coefficients are also in the ratio of 2:1:2. But these numbers are related to the number of things in a mole: the first and last numbers are two times Avogadro’s number, while the second number is Avogadro’s number. That means that the first and last numbers represent 2 mol, while the middle number is just 1 mol. Well, why not just use the number of moles in balancing the chemical equation?
2 H2 + O2 → 2 H2O
is the same balanced chemical equation we started with! What this means is that chemical equations are not just balanced in terms of molecules; they are also balanced in terms of moles. We can just as easily read this chemical equation as “two moles of hydrogen react with one mole of oxygen to make two moles of water.” All balanced chemical reactions are balanced in terms of moles.
Explanation:
Answer:
See explanation
Explanation:
Ga is in group 13 hence it must loose three electrons to form Ga^3+ in order to achieve the noble gas configuration because it has three electrons on its outermost shell.
O is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form O^2- since oxygen has six electrons on its outermost shell.
Br in group 17 has seven electrons in its outermost shell hence it must form Br^- (gain one electron) in order to attain the noble gas configuration.
P in group 15 must accept three electrons and form P^3- in order to attain the noble gas configuration since it has five electrons on its outermost shell.
S is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form S^2- since sulphur has six electrons on its outermost shell.
Mg in group 2 has two electrons on its outermost shell and must loose both to attain the noble gas configuration forming Mg^2+.
Al is in group 13 hence it must loose three electrons to form Al^3+ in order to achieve the noble gas configuration because it has three electrons on its outermost shell.
Se is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form Se^2- since selenium has six electrons on its outermost shell.
Lithium is in group 1 and must loose its only outermost electron in order to attain the noble gas configuration to form Li^+.
Rb is in group 1 and must loose its only outermost electron in order to attain the noble gas configuration to form Rb^+.
As in group 15 must accept three electrons and form As^3- in order to attain the noble gas configuration since it has five electrons on its outermost shell.
I in group 17 has seven electrons in its outermost shell hence it must form I^- (gain one electron) in order to attain the noble gas configuration.
Answer:
5.90 moles of ethane, C₂H₆, is burned in an excess of oxygen to produce 11.8 moles of CO₂.
Explanation:
To answer the question, we examine the chemical reaction as follows
2C₂H₆ ( g ) + 7O₂ ( g ) ⟶ 4CO₂ ( g ) + 6H₂O ( g )
From the chemical reaction it is seen that 2 moles of ethane, C₂H₆, is required to produce 4 moles of CO₂
This means that 1 mole of C₂H₆, is required to produce 2 moles of CO₂
Since we are asked to find out how many moles of CO 2 are produced when 5.90 moles of ethane is burned in an excess of oxygen and we already know that 1 mole of C₂H₆, is required to produce 2 moles of CO₂, we multiply both the 1 mole of C₂H₆ and the 2 moles of CO₂ by 5.90 to obtain;
(5.90 × 1 mole) of C₂H₆ is required to produce (5.90× 2 moles) of CO₂ or
5.90 moles of C₂H₆ is required to produce 11.8 moles of CO₂.
Answer:
Can you provide a picture?
Explanation:
Answer: 2 moles of 
will be formed.
Explanation:
To calculate the moles :
The balanced chemical equation is:
According to stoichiometry :
2 moles of 
give = 2 moles of
Thus 2 moles of give =
of
Thus 2 moles of will be formed.
