Answer:
C8H17N
Explanation:
Mass of the unknown compound = 5.024 mg
Mass of CO2 = 13.90 mg
Mass of H2O = 6.048 mg
Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:
For carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 13.90 = 3.791 mg
For hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 6.048 = 0.672 mg
For nitrogen, N:
Mass N = mass of unknown – (mass of C + mass of H)
Mass of N = 5.024 – (3.791 + 0.672)
Mass of N = 0.561 mg
Now, we can obtain the empirical formula for the compound as follow:
C = 3.791 mg
H = 0.672 mg
N = 0.561 mg
Divide each by their molar mass
C = 3.791 / 12 = 0.316
H = 0.672 / 1 = 0.672
N = 0.561 / 14 = 0.040
Divide by the smallest
C = 0.316 / 0.04 = 8
H = 0.672 / 0.04 = 17
N = 0.040 / 0.04 = 1
Therefore, the empirical formula for the compound is C8H17N
The amount of substance present in a certain object with a given half-life in terms of h can be expressed through the equation,
A(t) = (A(o))(0.5)^(t/h)
where A(t) is the amount of substance after t years and A(o) is the original amount. In this item we are given that A(t)/A(o) is equal to 0.89. Substituting the known values,
0.89 = (0.5)(t / 5730 years)
The value of t from the equation is 963.34 years.
<em>Answer: 963 years</em>
Answer:
d. Heat is released from the reaction
Explanation:
A negative enthalpy change indicates that it is an exothermic reaction. Exothermic reactions release heat.
Answer:
c) 22
Explanation:
Let's consider the following balanced equation.
N₂(g) + 3 H₂(g) ----> 2 NH₃(l)
According to the balanced equation, 34.0 g of NH₃ are produced by 1 mol of N₂. For 170 g of NH₃:
According to the balanced equation, 34.0 g of NH₃ are produced by 3 moles of H₂. For 170 g of NH₃:
The total gaseous moles before the reaction were 5.00 mol + 15.0 mol = 20.0 mol.
We can calculate the pressure (P) using the ideal gas equation.
P.V = n.R.T
where
V is the volume (50.0 L)
n is the number of moles (20.0 mol)
R is the ideal gas constant (0.08206atm.L/mol.K)
T is the absolute temperature (400.0 + 273.15 = 673.2K)
Step 1: Change density from g/mL to g/L;
0.807 g/mL = 807 g/L
Step 2: Find Moles of N₂;
As,
Density = Mass / Volume
Or,
Mass = Density × Volume
Putting Values,
Mass = 807 g/L × 1 L
Mass = 807 g
Also,
Moles = Mass / M.mass
Putting values,
Moles = 807 g / 28 g.mol⁻¹
Moles = 28.82 moles
Step 3: Apply Ideal Gas Equation to Find Volume of gas occupied,
As,
P V = n R T
V = n R T / P
Putting Values, remember! don't forget to change temperatue into Kelvin (25 °C + 273 = 298 K)
V = (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm
V = 704.76 L
