The new volume is 330.2 ml
<u><em>calculation</em></u>
The new volume is calculated using the Charles law formula
that is V1/T1= V2/T2
where T1= 25.0 c into kelvin = 25 +273 = 298 K
V1= 300.0 ml
T2 = 55.0 c into kelvin = 273 +55 =328 K
V2 = ? ml
make V2 the subject of the formula by multiplying both side by T2
V2= V1T2/ T1
V2 =[ (300.0 ml x 328 k) / 298 k} = 330.2 ml
Answer:
1. From water vapor to the dry ice;
2. The potential energy is higher before the water vapor condenses;
3. The thermal energy is higher in the 2.0 kg block.
Explanation:
1. The heat flows from the system with high temperature to the system with low temperature. The water vapor is at 298 K, and the dry ice is at 194.5 K.
2. The energy of the molecules is related to the temperature and the physics state. At the gas state, the molecules are more agitated, and the energy is higher than the liquid state. So, when the vapor condenses to a liquid, the energy decreases.
3. The thermal energy can be calculated by:
Q = m*c*ΔT
Where m is the mass, c is the specific heat, and ΔT the variation in the temperature. So, when the mass increase, thermal energy also increases.
Answer:
Mass = 5.33 g
Explanation:
Given data:
Mass of Al = 2.80 g
Mass of Cl₂ = 4.15 g
Theoretical yield of AlCl₃ = ?
Solution:
Chemical equation:
2Al + 3Cl₂ → 2AlCl₃
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 2.80 g/ 27 g/mol
Number of moles = 0.10 mol
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 4.15 g/71 g/mol
Number of moles = 0.06 mol
Now we will compare the moles of AlCl₃ with Al and Cl₂.
Cl₂ : AlCl₃
3 : 2
0.06 : 2/3×0.06 = 0.04
Al : AlCl₃
2 : 2
0.10 : 0.10
Number of moles of AlCl₃ produced by chlorine are less so it will be limiting reactant.
Mass of AlCl₃:Theoretical yield
Mass = number of moles ×molar mass
Mass = 0.04 mol × 133.34 g/mol
Mass = 5.33 g
Answer:
4.8 %
Explanation:
We are asked the concentration in % by mass, given the molarity of the solution and its density.
0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:
MW acetic acid = 60.0 g/mol
mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g
mass of solution = 1000 cm³ x 1.010 g/ cm³ (1l= 1000 cm³)
= 1010 g
% (by mass) = 48.00 g/ 1010 g x 100 = 4.8 %
Answer: 
Explanation:
cM 0 0
So dissociation constant will be:
Given: c = 0.15 M
pH = 1.86
= ?
Putting in the values we get:
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![1.86=-log[H^+]](https://tex.z-dn.net/?f=1.86%3D-log%5BH%5E%2B%5D)
![[H^+]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01)
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
As ![[H^+]=[ClCH_2COO^-]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BClCH_2COO%5E-%5D%3D0.01)
![K_a=1.67\times 10^{-3]](https://tex.z-dn.net/?f=K_a%3D1.67%5Ctimes%2010%5E%7B-3%5D)
Thus the vale of for the acid is
