Answer:
Molarity of HCl = 1.6M
Explanation:
The chemical reaction equation is;
HCl(aq) + NaOH(aq) —> NaCl(aq) + H2O(l)
Now, molarity = number of moles/volume
Thus, for NaOH, we have;
Number of moles = molarity × volume = 2M × (20/1000) L
Number of moles = 0.04 moles
Using the coefficients in the chemical equation above, we can find the corresponding number of moles for HCl.
Number of moles of HCl = 0.04 moles NaOH × (1 mole of HCl/1 mole of HCl) = 0.04 moles of HCl
Thus;
Molarity of HCl = 0.04/(25/1000)
Molarity of HCl = 1.6M
Answer:
It is none. Echo is formed when sound waves hit a hard surface.
Explanation:
Answer:
Mass of Fe produced = 44.7 g
Explanation:
Given data:
Mass of Fe produced in gram = ?
Mass of FeO react = 57 g
Solution:
Chemical equation:
2Al + 3FeO → 3Fe + Al₂O₃
Number of moles of FeO:
Number of moles = mass/ molar mass
Number of moles = 57 g/ 71.844 g/mol
Number of moles = 0.8 mol
Now we will compare the moles of Fe with FeO from balance chemical equation.
FeO : Fe
3 : 3
0.8 : 0.8
Mass of Fe:
Mass = Mass × molar mass
Mass = 0.8 mol × 55.845 g/mol
Mass = 44.7 g
Answer:
13.85 kJ/°C
-14.89 kJ/g
Explanation:
<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>
<em />
The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:
According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.
Qcomb + Qcal = 0
Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ
The heat capacity (C) of the calorimeter can be calculated using the following expression.
Qcal = C . ΔT
where,
ΔT is the change in the temperature
Qcal = C . ΔT
50.72 kJ = C . 3.661 °C
C = 13.85 kJ/°C
<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>
Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ
The heat of combustion per gram of B is:
Answer:
T₂ = 259.84 K
T₂ = -13.31 °C
Explanation:
Given data:
Initial pressure = 700 mmHg
Initial temperature = 30.0°C (30+273.15 K = 303.15 K)
Final temperature = ?
Final pressure = 600 mmHg
Solution:
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
700 mmHg /303.15 K = 600 mmHg / T₂
T₂ = 600 mmHg × 303.15 K / 700 mmHg
T₂ =181890 mmHg.K /700 mmHg
T₂ = 259.84 K
Temperature in celsius
259.84 K - 273.15 = -13.31 °C
