Explanation:
We have to find the number of moles of N₂ that are present in a sample that has a volume of 40.0 L at STP.
STP means Standard Conditions of Temperature and Pressure. These conditions are 273.15 K and 1 atm. We know that 1 mol of N₂ will occupy 22.4 L. We can use that ratio to find the answer to our problem.
1 mol of N₂ = 22.4 L
moles of N₂ = 40.0 L * 1 mol/(22.4 L)
moles of N₂ = 1.79 mol
Answer: 1.79 moles of nitrogen are present.
Answer:
K, the rate constant = 9.73 × 10^(-1)/s
Explanation:
r = K × [A]^x × [B]^y
r = Rate = 1.07 × 10^(-1)/s
K = Rate constant
A and B = Concentration in mol/dm^-3
A = 0.44M
B = 0.11M
x = Order of reaction with respect to A = 0
y = Order of reaction with respect to B = 1
Solving, we get
r/([A]^x × [B]^y) = K
K = 1.07 × 10^(-1)/s/(0.44^0 × 0.11^1)= 0.9727
K = 0.9727
Answer:
The empirical formula is CH2O, and the molecular formula is some multiple of this
Explanation:
In 100 g of the unknown, there are 40.0⋅g12.011⋅g⋅mol−1 C; 6.7⋅g1.00794⋅g⋅mol−1 H; and 53.5⋅g16.00⋅g⋅mol−1 O.
We divide thru to get, C:H:O = 3.33:6.65:3.34. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of CH2O, i.e. near enough to WHOLE numbers. Now the molecular formula is always a multiple of the empirical formula; i.e. (EF)n=MF.So 60.0⋅g⋅mol−1=n×(12.011+2×1.00794+16.00)g⋅mol−1.Clearly n=2, and the molecular formula is 2×(CH2O) = CxHyOz.
A pair of elements will most likely form an ionic bond if one is a metal and one is a nonmetal. These types of ionic compounds are composed of monatomic cations and anions. ( K, Cl)...