<span>2.28 kg x 213 kJ/kg = 486 kJ = 4.86E+05 J</span>
D) They both look uniform (the same) throughout.
<h3>Further explanation</h3>
Pure substance can be any element or compound and is formed from one type of atom/molecule only
Meanwhile, the solution is included in a mixture consisting of 2 or more pure substance
Pure substance can be formed through a chemical process while the mixture is through a physical process
Mixture can be separated by physical processes into components of pure substance while pure substance cannot
The mixture itself consists of a homogeneous and heterogeneous solution
The mixture can be divided into a homogeneous mixture if the composition/ratio of each substance in the mixture is the same and a heterogeneous mixture if the ratio of the composition of the substances is not the same (can be varied) in each place.
Mixtures can also be divided into solutions, suspensions, and colloids based mainly on the size of the particles
Homogeneous mixture = Solution
Heterogeneous mixture = suspension, and
The mixture is located between suspension and solution = Colloid
Answer:
m = 4450 g
Explanation:
Given data:
Amount of heat added = 4.45 Kcal ( 4.45 kcal ×1000 cal/ 1kcal = 4450 cal)
Initial temperature = 23.0°C
Final temperature = 57.8°C
Specific heat capacity of water = 1 cal/g.°C
Mass of water in gram = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 57.8°C - 23.0°C
ΔT = 34.8°C
4450 cal = m × 1 cal/g.°C × 34.8°C
m = 4450 cal / 1 cal/g
m = 4450 g
Answer:
Explanation:
Hello,
For the given chemical reaction:
We first must identify the limiting reactant by computing the reacting moles of Al2S3:
Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:
Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:
Finally, we compute the percent yield with the obtained 2.10 g:
Best regards.
