Answer:
65.2L
Explanation:
Using the general gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (Litres)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (Kelvin)
According to the information provided in this question,
P = 1.631 atm
V = ?
n = 4.3 moles
T = 28°C = 28 + 273 = 301K
Using PV = nRT
V = nRT/P
V = 4.3 × 0.0821 × 301 ÷ 1.631
V = 106.26 ÷ 1.631
V = 65.15
Volume of the gas = 65.2L
P1V1/T1=P2V2/T2 P2=1/2 P1 V1=1 T1=298K
1 P1/298= (1/2) P1V2/373 cross P1
1/298=1/2V2/373
1/298=1/V2 746
v2=746/298
V2=2.5L
An anchoring phenomenon anchors all of the learning within a unit. So, it is a unit level event that the classroom is trying to make sense of as they engage in a series of lessons.
Since the questions the students ask about the anchor drive the learning within the unit, the anchor should be complex and require an understanding of several big science ideas to explain.
At strategic moments, the class revisits the anchoring phenomenon to review their initial questions to see which they have answered, which they are making progress on, and what new questions they may have to help us continue learning about the phenomenon.
Throughout the unit, the classroom and each student should be given opportunities to share their thinking and how it relates to the anchoring phenomenon.
YOU SHOULD PUT IT IN YOUR OWN WORDS THOUGH <3
The anwser is atoms are destroyed
Answer:
Explanation:
Given parameters :
Volume of solution = 100mL
Absorbance of solution = 0.30
Unknown:
Concentration of CuSO₄ in the solution = ?
Solution:
There is relationship between the absorbance and concentration of a solution. They are directly proportional to one another.
A graph of absorbance against concentration gives a value of 0.15M at an absorbance of 0.30.
The concentration is 0.15M
Also, we can use: Beer-Lambert's law;
A = ε mC l
where εm is the molar extinction coefficient
C is the concentration
l is the path length
Since the εm is not given and assuming path length is 1;
Then we solve for the concentration.
