5 grams of NaHCO3 was added to 50 cm3 of 1.5 mol dm-3 hydrochloric acid, causing the temperature to decrease by 6.8 K. What is t
2 answers:
The balanced chemical equation for the neutralization of HCl with is:
Given weight of = 5g
Moles of =
Volume of HCl solution =
Assuming the density of the solution is 1.0 g/mL
Mass of HCl solution = 50 g
Overall mass of the solution = 50 g + 5 g = 55 g
To find the heat of neutralization, we calculate:
Q = m C ΔT
where m equals the mass of the solution = 55 g
C represents the specific heat capacity of the solution = 4.184
ΔT signifies the temperature change = 6.8 K = (6.8 - 273) C = -266.2
The enthalpy of neutralization per mole of
=
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Answer:
There are 5.5668 moles of water for every mole of CuSO₄.
Explanation:
The mass of anhydrous CuSO₄ is:
23.403g - 22.652g = 0.751g.
mass of crucible + lid + CuSO₄ - mass of crucible + lid
Given that the molar mass of CuSO₄ is 159.609g/mol, we calculate the moles:
0.751g × = 4.7052x10⁻³ moles CuSO₄
The mass of water in the initial sample is:
23.875g - 0.751g - 22.652g = 0.472g.
mass of crucible + lid + CuSO₄ hydrate - CuSO₄ - mass of crucible + lid
As the molar mass of H₂O is 18.02g/mol, we find the moles:
0.472g × = 2.6193x10⁻² moles H₂O
The mole ratio of H₂O to CuSO₄ is:
2.6193x10⁻² moles H₂O / 4.7052x10⁻³ moles CuSO₄ = 5.5668
This indicates there are 5.5668 moles of water per mole of CuSO₄.
I hope this is helpful!
Answer:
Complete Question:
Equimolar quantities of CH3OH(l) and C2H5OH(l) are placed in separate 2.0 L containers that have been evacuated beforehand. Pressure gauges are attached to each container, and the temperature is maintained at 300 K. In both containers, liquid is consistently visible at the bottom. The varying pressure within the vessel that contains CH3OH(l) is illustrated below.
In comparison to the equilibrium vapor pressure of CH3OH(l) at 300 K, the equilibrium vapor pressure of C2H5OH(l) at 300 K is
ANSWER : lower, since the London dispersion forces among C2H5OH molecules surpass those among CH3OH molecules.
Explanation:
To clarify the answer provided, let’s begin by defining some concepts.
The London dispersion force is the least strong type of intermolecular force. It is a temporary force that arises when the electron arrangement in two neighboring atoms creates transient dipoles.
The vapor pressure of a liquid reflects the equilibrium pressure of its vapor above the liquid (or solid); specifically, it represents the pressure associated with the evaporation of a liquid (or solid) in a sealed environment above the substance.
The pressure will be lower due to the stronger London dispersion forces acting between C2H5OH molecules compared to those between CH3OH molecules. This implies that when intermolecular forces are stronger, they intensify the interactions binding the substance together, thereby reducing the liquid's vapor pressure at any given temperature and making it more difficult to vaporize the substance.
Note: The London dispersion force for C2H5OH is more substantial than for CH3OH because C2H5OH has more electrons than CH3OH.
Response:
Here's my calculation
Clarification:
Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We need to determine the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's compile all the information in one location.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
3. Plot the initial values
The graph below visualizes the initial concentrations as plotted on the vertical axis.
