3.00 g
5.56 g
8.40 g
Answer : The amount left of leutium-176 will be, 2.10 g
Solution :
First we have to calculate the rate constant, we use the formula :
Now we have to calculate the amount left of the sample.
Expression for rate law for first order kinetics is given by :
where,
k = rate constant =
t = decay time =
a = initial amount of the sample = 16.8 g
a - x = amount left after decay process = ?
Now put all the given values in above equation, we get
Therefore, the amount left of leutium-176 will be, 2.10 g
Answer:
A.
Explanation:
B. Reflection
C. Diffraction
D. Resonance
b) [Ha} < [A-]
c) [HA]=[A-]
Answer:
Light as a wave: Light can be described (modeled) as an electromagnetic wave. In this model, a changing electric field creates a changing magnetic field. This changing magnetic field then creates a changing electric field and BOOM - you have light. ... So, Maxwell's equations do say that light is a wave.
Explanation:
Hope this helps
Answer:
Light as a wave: Light can be described (modeled) as an electromagnetic wave. In this model, a changing electric field creates a changing magnetic field. This changing magnetic field then creates a changing electric field and BOOM - you have light. ... So, Maxwell's equations do say that light is a wave.
Explanation:
Credit for 2021FOLLOWme
The enthalpy change (ΔH) for the neutralization of 0.1 moles of 1.0 M NaOH with 0.1 moles of 1.0 M HCl in a coffee-cup calorimeter is approximately 28.05 kJ/mol.
To calculate the enthalpy change (ΔH) for the neutralization of HCl by NaOH, you can use the equation:
ΔH = q / moles of limiting reactant
First, let's find the moles of the reactants. We have 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl. Since we know the volumes and concentrations, you can find the moles of each reactant using the formula:
moles = (volume in L) × (concentration in mol/L)
For NaOH:
moles of NaOH = (100.0 mL / 1000 mL/L) × 1.0 mol/L = 0.1 moles
For HCl:
moles of HCl = (100.0 mL / 1000 mL/L) × 1.0 mol/L = 0.1 moles
Now, you need to determine the limiting reactant. The balanced chemical equation for the neutralization of HCl by NaOH is:
NaOH + HCl → NaCl + H₂O
The stoichiometric ratio of NaOH to HCl is 1:1, which means they react in a 1:1 ratio. Since both reactants have 0.1 moles, neither is in excess. Therefore, the reactant that limits the reaction is the one that is present in the smaller amount, which is NaOH in this case.
Now, calculate the heat absorbed or released (q) using the equation:
q = mΔTC
Where:
m is the mass (in grams) of the solution, which we can calculate using the density of 1.0 g/cm³ and the volume (in mL).
ΔT is the change in temperature.
C is the specific heat capacity (given as 4.18 J/g°C).
For the volume of 100.0 mL, the mass is 100.0 g (since 100.0 mL = 100.0 g, given the density is 1.0 g/cm³).
ΔT = Final temperature - Initial temperature
ΔT = 31.38°C - 24.68°C = 6.70°C
Now, calculate q for the reaction:
q = 100.0 g × 6.70°C × 4.18 J/g°C = 2804.76 J
Finally, calculate the enthalpy change (ΔH) by dividing q by the moles of the limiting reactant:
ΔH = 2804.76 J / 0.1 moles = 28047.6 J/mol
Since the enthalpy change is typically expressed in kJ/mol, divide by 1000 to convert J to kJ:
ΔH = 28.05 kJ/mol
So, the enthalpy change for the neutralization of HCl by NaOH is approximately 28.05 kJ/mol.
To know more about moles:
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