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andreev551 [17]
3 years ago
8

A graduated cylinder contains 10.00 mL water. A 14.74 g piece of aluminum is added to the water, and the volume rises to 15.46 m

L. What is the density of the aluminum in g/mL
Chemistry
1 answer:
Lyrx [107]3 years ago
8 0

Answer:

2.67 g/mL

Explanation:

Initial Volume of water = 10.0 mL

Final Volume pf water = 15.46 mL

Mass = 14.74g

Density = Mass / Volume of aluminium

Volume of aluminium = Final Volume of water - Initial Volume of water = 15.46 - 10 = 5.46mL

Density = 14.74 / 5.46 = 2.67 g/mL

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400 mL of gas is contained at 300 mmHg and 0 °C. What will its volume be at 140 mmHg and 100 °C? 0°C 100°C
PIT_PIT [208]

Answer:

1171.12 mL

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

V1 = initial volume (milliliters)

V2 = final volume (milliliters)

T1 = initial temperature (Kelvin)

T2 = final temperature (Kelvin)

According to the information provided in this question:

P1 = 300 mmHg

P2 = 140 mmHg

V1 = 400 mL

V2 = ?

T1 = 0°C = 273K

T2 = 100°C = 100 + 273 = 373K

Using P1V1/T1 = P2V2/T2

300 × 400/273 = 140 × V2/373

120000/273 = 140V2/373

120000 × 373 = 273 × 140V2

44760000 = 38220V2

V2 = 44760000 ÷ 38220

V2 = 1171.115

The new volume is 1171.12 mL

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3 years ago
Which one of these is not an example of intercellular communication?
Mashcka [7]

the answer is B. ribosomes make protien

7 0
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What is the volume in liters of 321 g of liquid with a density of 0.84 g/mL
ss7ja [257]
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The normal freezing point of a certain liquid
slavikrds [6]

Answer : The molal freezing point depression constant of liquid X is, 4.12^oC/m

Explanation :  Given,

Mass of urea (solute) = 5.90 g

Mass of liquid X (solvent) = 450 g  = 0.450 kg

Molar mass of urea = 60 g/mole

Formula used :  

\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of liquid X Kg}}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution = -0.5^oC

\Delta T^o = freezing point of liquid X = 0.4^oC

i = Van't Hoff factor = 1 (for non-electrolyte)

K_f = Molal-freezing-point-depression constant = ?

m = molality

Now put all the given values in this formula, we get

0.4^oC-(-0.5^oC)=1\times K_f\times \frac{5.90g}{60g/mol\times 0.450kg}

K_f=4.12^oC/m

Therefore, the molal freezing point depression constant of liquid X is, 4.12^oC/m

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3 years ago
What was Johann Dobereiner’s contribution to the development of the periodic table?
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<span>he introduced his law of triads. each triad was a group of three elements</span>
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