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
Answer : The dipole-dipole interaction.
Explanation : The kind of bond that is created by a weak electrical attraction between two polar molecules is dipole-dipole interaction.
The type of force which occurs between the positive end of one molecule and the negative end of another molecule: is dipole-dipole interaction.
For better understanding, please refer the attachment.
1 gram of sugar because super molecules are bigger then the ions of dissolved salt
The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 1.25 mL
<u>Explanation:</u>
To calculate the volume of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of stomach acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of HCl neutralized is 1.25 mL
Answer:
%yield of NH₃ = 30%
Explanation:
Actual yield of NH₃ = 40.8g
Theoretical yield = ?
Equation of reaction
N₂ + 3H₂ → 2NH₃
Molar mass of NH₃ = 17g/mol
Molarmass of N = 14.00
2 molecules of N = 2 * 14.00 = 28g/mol
Number of moles = mass / molar mass
Mass = number of moles * molar mass
Mass = 1 * 28.00 = 28g of N₂ (the number of moles of N₂ from the equation is 1).
From the equation of reaction,
28g of N₂ produce (2 * 17)g of NH₃
28g of N₂ = 34g of NH₃
112g of N₂ = x g of NH₃
X = (112 * 34) / 28
X = 136g of NH₃
Theoretical yield = 136g of NH₃
% yield = (actual yield / theoretical yield) * 100
% yield = (40.8 / 136) * 100
% yield = 0.3 * 100
% yield = 30%
