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Readme [11.4K]
2 years ago
6

Compare Xavier's cup design with the criteria and constraints given. What parts of Xavier's cup design were successful? What par

ts were unsuccessful?
Chemistry
1 answer:
Nata [24]2 years ago
6 0

Answer:

The lid is successful and so is the handle. I dont think the thick glass would work out as good unless you only keep inside and the compartment cylinder is what made thier beverages taste different from what I think. 

Explanation:

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dlinn [17]

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Explanation:

She didn't get the answer because she didn't add the them well , due to the bracket present.

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What does biodiversity mean?
kobusy [5.1K]

Answer:

Having as wide a range of organisms as possible.

Hope it helps! :)

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3 years ago
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Water is placed in a graduated cylinder and the volume is recorded as 43.5 mL. A homogeneous sample of metal pellets with a mass
Ipatiy [6.2K]

Answer:

1.8g

Explanation:

Initial volume = 43.5ml

Final volume = 49.4ml

Mass = 10.88g

Density = ?

Volume = Final volume - initial volume

= 49.4 - 43.5

= 5.9ml

Density = Mass/volume

Density = 10.88/5.9

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4 0
3 years ago
Which of the following is an example of oxidation? A. Charcoal is placed on a grill and lighting fluid poured over it. B. A car
Vikentia [17]

Answer:

<u>The Answer is (B)  A car get rusty over the course of few years</u>Explanation:

<u>Explanation:</u>

  • <u>Oxidation </u>refers to the process of loss of electrons by a molecule,atom or ion during a chemical reaction.The process which is just the opposite of oxidation is reduction,it occurs when their is gain of electrons .
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4 0
3 years ago
Which solute would be more effective at lowering the freezing point of water: MgCl2 and KNO3? Explain.
Phantasy [73]

Answer:

AlCl₃.

Explanation:

Adding solute to water causes depression of the boiling point.

The depression in freezing point (ΔTf) can be calculated using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kf is the molal depression constant of water.

m is the molality of the solution (m = 1.0 m, for all solutions).

(1) NaCl:

i for NaCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (NaCl) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(2) MgCl₂:

i for MgCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

∴ ΔTb for (MgCl₂) = i.Kb.m = (3)(Kf)(1.0 m) = 3(Kf).

(3) NaCl:

i for KBr = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (KBr) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(4) AlCl₃:

i for AlCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.

∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(1.0 m) = 4(Kf).

So, the ionic compound will lower the freezing point the most is: AlCl₃

4 0
3 years ago
Read 2 more answers
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