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Zigmanuir [339]
3 years ago
6

Mickey notices that his dog Pluto is suffering from fleas which causes his hair to fall out. His friend Goofy tells him that rub

bing cabbage on his hair is the perfect cure. Mickey decides to test this cure by rubbing Pluto with cabbage for one week. After a week of treatment, the fleas are gone and Pluto’s hair looks healthy and feels a lot better. what is conclusion
Chemistry
1 answer:
dsp733 years ago
4 0

Answer:

rubbing dogs with cabbage will get rid of fleas.

Explanation:

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How many liters of solution can be produced from 2.5 moles of solute if a 2.0 M
Firlakuza [10]

Answer:

The answer is 1.25 L

Explanation:

5 0
3 years ago
Which of the following would not be helpful when distinguishing between sugar and table salt?
inysia [295]
Color they look the same hope this helps.
4 0
3 years ago
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What mass of H2 is needed to react with 8.75 g of O2 according to the following equation: O2(g) + H2(g) → H2O(g)?
alina1380 [7]

Mass of H₂ needed to react with O₂ : 1.092 g

<h3>Further explanation</h3>

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

Reaction

O₂(g) + 2H₂(g) → 2H₂O(g)

mass of O₂ : 8.75 g

mol O₂(MW=32 g/mol) :

\tt \dfrac{8.75}{32}=0.273

From the equation, mol ratio of O₂ : H₂ = 1 : 2, so mol H₂ :

\tt \dfrac{2}{1}\times 0.273=0.546

Mass H₂ (MW=2 g/mol) :

\tt 0.546\times 2=1.092~g

4 0
3 years ago
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How well did tossing the pennies simulate half lives?
SSSSS [86.1K]
Its a 50% chance that approx 1/2 of the pennies will land on tails. The next toss will result the same. and so on and so on. showing how a reaction would slowly eliminate 1/2 of the remaining lives per reaction, until nothing is left. I think it is a good stimulation.
5 0
3 years ago
A certain gas at 2oC and 1.00 atm pressure fills a 4.0 container. What volume will the gas occupy at 100oC and 780 torr pressure
artcher [175]

The final volume V₂=4.962 L

<h3>Further explanation</h3>

Given

T₁=20 + 273 = 293 K

P₁= 1 atm

V₁ = 4 L

T₂=100+273 = 373 K

P₂=780 torr=1,02632 atm

Required

The final volume

Solution

Combined gas law :

P₁V₁/T₁=P₂V₂/T₂

Input the value :

V₂=(P₁V₁T₂)/(P₂T₁)

V₂=(1 x 4 x 373)/(1.02632 x 293)

V₂=4.962 L

6 0
3 years ago
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