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dem82 [27]
3 years ago
5

Simple question ill name u brainiest

Chemistry
2 answers:
BigorU [14]3 years ago
4 0

Answer:

Test tube :)

Explanation:

Yuki888 [10]3 years ago
4 0

Im pretty shure its D- graduated cylendar

A flask holds much more than that a test tube doesnt have mesure meants (i dont tink so at least?) and a beaker also much more and is harder to explain..

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Considering all types of aerosols, what is their net effect on the climate system:
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3 years ago
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The most prominent topographical feature of the ocean discovered in the 1960's was _____.
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The moutain ridge which circles the earth.
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A solution is prepared from 4.5701 g of magnesium chloride and 43.238 g of water. The vapor pressure of water above this solutio
balandron [24]

Answer:

i = 2.483

Explanation:

The vapour pressure lowering formula is:

Pₐ = Xₐ×P⁰ₐ <em>(1)</em>

For electrolytes:

Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ

Where:

Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)

4.5701g of MgCl₂ are:

4.5701g ₓ (1mol / 95.211g) = 0.048000 moles

43.238g of water are:

43.238g ₓ (1mol / 18.015g) = 2.400 moles

Replacing in (1):

0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm

0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)

2.4mol + i*0.048mol = 2.4mol / 0.9527

2.4mol + i*0.048mol = 2.5192mol

i*0.048mol = 2.5192mol - 2.4mol

i = 0.1192mol / 0.048mol

<em>i = 2.483</em>

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I hope it helps!

4 0
3 years ago
Nuclear Processes Unit Test
Finger [1]

Answer:

?

Explanation:

8 0
3 years ago
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How much heat energy would be needed to raise the temperature of a 223 g sample of aluminum [(C=0.895 Jig Cy from 22.5°C to 55 0
dsp73

Answer : The heat energy needed would be, 6486.5125 J

Explanation :

To calculate the change in temperature, we use the equation:

q=mc\Delta T\\\\q=mc(T_2-T_1)

where,

q = heat needed = ?

m = mass of aluminum = 223 g

c = specific heat capacity of aluminum = 0.895J/g^oC

\Delta T = change in temperature

T_1 = initial temperature = 22.5^oC

T_2 = final temperature = 55.0^oC

Putting values in above equation, we get:

q=223g\times 0.895J/g^oC\times (55.0-22.5)^oC

q=6486.5125J

Therefore, the heat energy needed would be, 6486.5125 J

5 0
3 years ago
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