Answer: The coefficient of nitrogen in the given equation is 2.
Explanation: The reaction for the oxidation of methamphentamine with oxygen gas in the body is given by:
By Stoichiometry,
4 moles of methamphentamine reacts with 55 moles of oxygen gas to produce 40 moles of carbon dioxide gas, 30 moles of water and 2 moles of nitrogen gas.
Coefficient of 
Coefficient of 
Coefficient of 
Coefficient of 
Coefficient of 
Hence, the coefficient of nitrogen in the given equation is 2.
Kiloliters and megaliters
Answer:
C₄H₂N₂
Explanation:
First we<u> calculate the moles of the gas</u>, using PV=nRT:
P = 2670 torr ⇒ 2670/760 = 3.51 atm
V = 300 mL ⇒ 300/1000 = 0.3 L
T = 228 °C ⇒ 228 + 273.16 = 501.16 K
- 3.51 atm * 0.3 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 501.16 K
Now we<u> calculate the molar mass of the compound</u>:
- 2.00 g / 0.0256 mol = 78 g/mol
Finally we use the percentages given to<em> </em><u>calculate the empirical formula</u>:
- C ⇒ 78 g/mol * 61.5/100 ÷ 12g/mol = 4
- H ⇒ 78 g/mol * 2.56/100 ÷ 1g/mol = 2
- N ⇒ 78 g/mol * 35.9/100 ÷ 14g/mol = 2
So the empirical formula is C₄H₂N₂
Answer:
B. the products have a smaller number of available energy microstates than the reactants.
<span>Ca3(PO4)2 + 3 H2SO4 = 3 CaSO4 + 2 H3PO4</span>
<span>Reaction type: double replacement
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