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Alinara [238K]
2 years ago
9

A solid cylinder has a radius of 2cm and a length of 7cm.It has a density of 3.1g/cm3.What is the mass of the cylinder?

Chemistry
1 answer:
Harman [31]2 years ago
4 0
272.69 grams

This is the answer

I hope this help
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In the titration of HCl with NaOH, the equivalence point is determined
kondaur [170]

Answer:

In the titration of HCl with NaOH, the equivalence point is determined from the point where the phenolphthalein turns pink and then remains pink on swirling.

Explanation:

The equivalence point is the point at which exactly enough titrant (NaOH) has been added to react with all of the analyte (HCl). Up to the equivalence point, the solution will be acidic because excess HCl remains in the flask.

Phenolphtalein is chosen because it changes color in a pH range between 8.3 – 10. Phenolphthalein is naturally colorless but turns pink in alkaline solutions. It remains colorless throughout the range of acidic pH levels, but it begins to turn pink at a pH level of 8.3 and continues to a bright purple in stronger alkalines.

It will appear pink in basic solutions and clear in acidic solutions.

The more NaOH added, the more pink it will be. (Until pH≈ 10)

In strongly basic solutions, phenolphthalein is converted to its In(OH)3− form, and its pink color undergoes a rather slow fading reaction and becomes completely colorless above 13.0 pH

a. from the point where the pink phenolphthalein turns colorless and then remains colorless on swirling.

⇒ the more colorless it turns, the more acid the solution. (More HCl than NaOH)

b. from the point where the phenolphthalein turns pink and then remains pink on swirling.

The equivalence point is the point where phenolphtalein turns pink and remains pink ( Between ph 8.3 and 10). (

Although, when there is hydrogen ions are in excess, the solution remains colorless. This begins slowely after ph= 10 and can be noticed around ph = 12-13

c. from the point where the pink phenolphthalein first turns colorless and then the pink reappears on swirling.

Phenolphthalein is colorless in acid solutions (HCl), and will only turn pink when adding a base like NaOH

d. from the point where the colorless phenolphthalein first turns pink and then disappears on swirling

Phenolphthalein is colorless in acid or neutral solutions. Once adding NaOH, the solution will turn pink. The point where the solution turns pink, and stays pink after swirling is called the equivalence point. When the pink color disappears on swirling, it means it's close to the equivalence point but not yet.

3 0
2 years ago
In multicellular organismscells that become specialized to perform a specific function are called?
Zolol [24]
The cell proliferates to produce many cells that result in multicellar organism.

answer
8 0
2 years ago
The diagram shows Earth's position in its orbit around the Sun at the beginning of each season
alexdok [17]

Answer: the answer is option 3. C

Explanation:

8 0
2 years ago
This means that if there are 15.0 g of starting chemicals, also known as
Serggg [28]

Answer:

Reactants

Explanation:

"Starting chemicals", the substances present before a reaction occurs, are called reactants.

The results of the reaction are called products, which you would also have 15.0g.

4 0
3 years ago
Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica
Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = d_{ideal} = \dfrac{P\times M}{RT}

here:

P = pressure of the O₂ gas = 310 bar

= 310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂  gas = 32 g/mol

∴

d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}

d_{ideal} = 287.37 g/L

To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol²    &

b = 0.0319  L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}

310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}

310V^3 -44.389V^2+1.382V-0.044=0

After solving;

V = 0.1152 L

∴

d_{Van \ der \ Waal} = \dfrac{32}{0.1152}

d_{Van \ der \ Waal} = 277.77  g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

5 0
3 years ago
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