Answer:
what each cells job is divided
Explanation:
Answer:
53.6 g of N₂H₄
Explanation:
The begining is in the reaction:
N₂(g) + 2H₂(g) → N₂H₄(l)
We determine the moles of each reactant:
59.20 g / 28.01 g/mol = 2.11 moles of nitrogen
6.750 g / 2.016 g/mol = 3.35 moles of H₂
1 mol of N₂ react to 2 moles of H₂
Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.
2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine
Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄
Let's convert the moles to mass:
1.67 mol . 32.05 g/mol = 53.6 g
Let's eliminate these one by one.
The first pair would not be the same, as X would most likely be in group IA, and Y would be in group VIIA, because of their tendency to gain and lose electrons.
The second pair would also violate the same rule, but X would most likely be in group IIA, and Y would most likely be in group VIA.
The third pair would not be the same, as X is most likely in group VIIA, and since Y has eight valence electrons, it is most likely a noble gas.
The final pair has X with atomic number 15, making it phosphorous. Phosphorous wants to gain 3 electrons to have a full octet of 8 outer "valence" electrons, and Y would also like to gain 3 electrons. This means it is possible that the final pair would be in the same group.
Answer:
Fe₃Si₇
Explanation:
In order to determine the empirical formula, we have to follow a series of steps.
Step 1: Determine the percent composition
Fe: 46.01%
Si: 53.99%
Step 2: Divide each percentage by the atomic mass of the element
Fe: 46.01/55.85 = 0.8238
Si: 53.99/28.09 = 1.922
Step 3: Divide all the numbers by the smallest one
Fe: 0.8238/0.8238 = 1
Si: 1.922/0.8238 = 2.33
Step 4: Multiply by numbers that make the coefficients whole.
Fe: 1 × 3 = 3
Si: 2.33 × 3 = 7
The empirical formula is Fe₃Si₇.
Answer: d) -705.55 kJ
Explanation:
Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.
Reversing the reaction, changes the sign of 
On multiplying the reaction by 
, enthalpy gets half:
Thus the enthalpy change for the given reaction is -705.55kJ
