Answer:
The system gains 126100 J
Explanation:
The heat can be calculated by the equation:
Q = nxCxΔT, where Q is the heat, C is the heat capacity,n is the number of moles and ΔT is the variation of temperature (final - initial). The number of moles is the mass divided by the molar mass, so:
n = 250/4 = 62.5 mol.
The system must be in thermal equilibrium with the surroundings, so if the temperature of the surroundings decreased 97 K, the temperature of the system increased by 97 K, so ΔT = 97 K
Q = 62.5x20.8x97
Q = 126100 J
<span>One degree Celsius indicates the same temperature change as one </span>
Conservation of mass is the underlying principle of balancing equation. When we balance equation, this means that we acknowledge that before and after the chemical reaction, the elements are conserved. To balance the chemical equation, we add coefficients before each reactant and product. Here are the following answers:
Reaction 1:
<span>2Al
3ZnCl2
3Zn
2AlCl3
Reaction 2:
</span><span>4NH3
3O2
2N2
6H2O</span>
Answer:
Explanation:
<u>1) Chemical equation (given)</u>
<u>2) Theoretical yield</u>
<u>a) Convert mass of NaHCO₃ to moles:</u>
- n = mass in grams / molar mass
- molar mass = 84.007 g/mol
- n = 2.36 g / 84.007 g/mol = 0.02809 mol
<u>b) Mole ratio:</u>
- 2 mol NaHCO₃ : 1 mol H₂CO₃
<u>c) Proportionality:</u>
- 2 mol NaHCO₃ / mol H₂CO₃ = 0.02809 mol NaHCO₃ / x
⇒ x = 0.2809 / 2 mol H₂CO₃ = 0.01405 mol H₂CO₃
<u>3) Actual yield</u>
<u>a) Mass balance</u>: 2.36 g - 1.57 g = 0.79 g
<u>b) Convert 0.79 g of carbonic acid to number of moles</u>:
- n = mass in grams / molar mass
- n = 0.79 g / 62.03 g/mol = 0.01274 mol
<u>4) Percentage yield, y (%)</u>
- y (%) = actual yield / theoretical yield × 100
- y (%) = 0.1274 mol / 0.1405 mol × 100 = 90.68%
The answer must show 3 significant figures, so y(%) = 90.7%.
Answer:
96%
Explanation:
Step 1: Write the balanced neutralization reaction
Al(OH)₃ + 3 HCl ⇒ AlCl₃ + 3 H₂O
Step 2: Calculate the theoretical yield of AlCl₃
According to the balanced equation, the mass ratio of Al(OH)₃ to AlCl₃ is 81.03:133.34.
28 g Al(OH)₃ × 133.34 g AlCl₃/81.03 g Al(OH)₃ = 46 g AlCl₃
Step 3: Calculate the percent yield of AlCl₃
The real yield of AlCl₃ is 44 g. We can calculate the percent yield using the following expression.
%yield = real yield / theoretical yield × 100%
%yield = 44 g / 46 g × 100% = 96%