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baherus [9]
3 years ago
7

Why are chromosomes not always visible?

Chemistry
1 answer:
Mashcka [7]3 years ago
5 0

Answer:

why are chromosomes not always visible? I'm going with a

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For an experiment, you place 15 g of ice with a temperature of -10 C into a cup and label
melamori03 [73]
The correct answer is c. Temperature is the average kinetic energy of a sample so if two samples have the same temperature they will also have the same average kinetic energy. I hope this helps. Let me know if anything is unclear.
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Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /
barxatty [35]

Answer:

[Pb^{2+}]=3.9 \times 10^{-2}M

this is the concentration required to initiate precipitation

Explanation:

PbCl_2  ⇄ Pb^{2+}+2Cl^-

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

Ip=[Pb^{2}][2Cl^-]^2=Ksp

Ksp=2.4\times 10^{-4}

lets solubility=S

[Pb^{2+}] = S

[Cl^-]=2S

Ksp=[Pb^{2+}]\times [Cl^-]^2

Ksp=S \times (2S)^2

Ksp=4S^3

S=\sqrt[3]{\frac{Ksp}{4} }

S=3.9\times 10^{-2}

[Pb^{2+}]=3.9 \times 10^{-2}M this is the concentration required to initiate precipitation

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3 years ago
!!!!!!!!!!!!SCIENCE!!!!!!!!!!!!<br><br> How are radio waves used on Earth?
nika2105 [10]

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How much heat is absorbed when a 298.3 g piece of brass goes from 30.0 to 150
igor_vitrenko [27]

Answer:

Q = 1360.248 j

Explanation:

Given data:

Mass of brass = 298.3 g

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Final temperature = 150°C

Specific heat capacity of brass = 0.038 J/g.°C

Heat absorbed = ?

SOLUTION:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

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ΔT = 150°C - 30.0°C  

ΔT = 120°C

Q = 298.3 g × 0.038 J/g.°C × 120°C

Q = 1360.248 j

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3 years ago
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Answer

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