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dolphi86 [110]
3 years ago
15

Is mud a homogeneous or heterogeneous mixture?

Chemistry
1 answer:
madreJ [45]3 years ago
3 0
Heterogeneous mixture
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ololo11 [35]

Answer:

hi here goes your answer

Explanation:

iv. The lower the PH, the weaker the base

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3 years ago
Explain the Anchoring Phenomenon (summary)
Virty [35]
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
3 0
3 years ago
Part 1: What is the final volume in milliliters when 0.730 L of a 44.8 % (m/v) solution is diluted to 23.3 % (m/v)?
Andre45 [30]

part 1 : the final volume : 1.404 L

part 2 : the initial concentration : 4.06 M

<h3>Further explanation </h3>

Dilution is the process of adding a solvent to get a more dilute solution.

The moles(n) before and after dilution are the same.

Can be formulated :

M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution  

V₁ = volume of the solution before dilution  

M₂ = Molarity of the solution after dilution  

V₂ = Molarity volume of the solution after dilution

part 1 :

M₁=44.8%

V₁=0.73 L

M₂=23.3%

\tt V_2=\dfrac{M_1.V_1}{M_2}\\\\V_2=\dfrac{44.8\times 0.73}{23.3}\\\\V_2=1.404~L

part 2 :

V₁=739 ml=0.739 L

V₂=1.5 L

M₂=2

\tt M_1=\dfrac{M_2.V_2}{V_1}\\\\M_1=\dfrac{2\times 1.5}{0.739}\\\\M_1=4.06

6 0
3 years ago
Identify the mixture of powdered charcoal and powdered sugar and suggest a technique for separating their components
mihalych1998 [28]
A- Identify the mixture:
The mixture of powdered charcoal and powdered sugar is considered as a homogeneous mixture. This means that you cannot identify the components with naked eye as they are uniformly distributed in the mixture.

B- Separate components:
You ca separate the charcoal powder from the sugar powder using the following steps:
1- add water. Sugar will dissolve in water while charcoal won't.
2- filter the solution where the powdered charcoal will remain on the filter paper and the solution of powder will pass through.
3- boil the sugar solution (above 100 degrees celcius). The water will evaporate and the sugar will precipitate.
5 0
3 years ago
Can somebody help me please
LenaWriter [7]

Answer:

0.1139952 is the answer

Explanation:

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