Answer:
d) cut the large sized Cu solid into smaller sized pieces
Explanation:
The aim of the question is to select the right condition for that would increases the rate of the reaction.
a) use a large sized piece of the solid Cu
This option is wrong. Reducing the surface area decreases the reaction rate.
b) lower the initial temperature below 25 °C for the liquid reactant, HNO3
Hugher temperatures leads to faster reactions hence this option is wrong.
c) use a 0.5 M HNO3 instead of 2.0 M HNO3
Higher concentration leads to increased rate of reaction. Hence this option is wrong.
d) cut the large sized Cu solid into smaller sized pieces
This leads to an increased surface area of the reactants, which leads to an increased rate of the reaction. This is the correct option.
since the concentration of Carbon Dioxide will increase, it would make Q > K, cause equilibrium to shift in the direction with less moles of gas to alleviate the extra pressure. In this case, the reaction will shift left because there are fewer moles of gas present.
<h3>
Answer:</h3>
0.387 J/g°C
<h3>
Explanation:</h3>
- To calculate the amount of heat absorbed or released by a substance we need to know its mass, change in temperature and its specific heat capacity.
- Then to get quantity of heat absorbed or lost we multiply mass by specific heat capacity and change in temperature.
- That is, Q = mcΔT
in our question we are given;
Mass of copper, m as 95.4 g
Initial temperature = 25 °C
Final temperature = 48 °C
Thus, change in temperature, ΔT = 23°C
Quantity of heat absorbed, Q as 849 J
We are required to calculate the specific heat capacity of copper
Rearranging the formula we get
c = Q ÷ mΔT
Therefore,
Specific heat capacity, c = 849 J ÷ (95.4 g × 23°C)
= 0.3869 J/g°C
= 0.387 J/g°C
Therefore, the specific heat capacity of copper is 0.387 J/g°C
Answer:
0.007 mol
Explanation:
We can solve this problem using the ideal gas law:
PV = nRT
where P is the total pressure, V is the volume, R the gas constant, T is the temperature and n is the number of moles we are seeking.
Keep in mind that when we collect a gas over water we have to correct for the vapor pressure of water at the temperature in the experiment.
Ptotal = PH₂O + PO₂ ⇒ PO₂ = Ptotal - PH₂O
Since R constant has unit of Latm/Kmol we have to convert to the proper unit the volume and temperature.
P H₂O = 23.8 mmHg x 1 atm/760 mmHg = 0.031 atm
V = 1750 mL x 1 L/ 1000 mL = 0.175 L
T = (25 + 273) K = 298 K
PO₂ = 1 atm - 0.031 atm = 0.969 atm
n = PV/RT = 0.969 atm x 0.1750 L / (0.08205 Latm/Kmol x 298 K)
n = 0.007 mol
