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3 years ago

What is the base metric unit for measuring volume?

1 answer:

5 0

I got this idk if it's correct: For measuring large distances, the kilometer (103 or 1000 meters) is often used. The basic unit of volume in the metric system is the liter (l). The most common derived unit is the milliliter (ml) (10-3 or 1/1000 of a liter). The volume of a milliliter is equal to the volume of a cube 1 centimeter per side.

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An aqueous solution of calcium hydroxide is standardized by titration with a 0.120 M solution of hydrobromic acid. If 16.5 mL of

Artist 52 [7]

<u>Answer:</u> The molarity of calcium hydroxide in the solution is 0.1 M

<u>Explanation:</u>

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HBr$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.120M\\V_1=27.5mL\\n_2=2\\M_2=?M\\V_2=16.5mL$

Putting values in above equation, we get:

$1\times 0.120\times 27.5=2\times M_2\times 16.5\\\\M_2=0.1M$

Hence, the molarity of $Ca(OH)_2$ in the solution is 0.1 M.

7 0

2 years ago

Will sodium iodide react with bromine to produce sodium bromide and iodine? Why or why not?

Gre4nikov [31]

Answer:

Explanation:

The higher the period the higher the activity of an element, therefore, since iodine is in period 6 and bromine is in period 5, the described reaction is not possible due to the fact that bromine is less active

4 0

1 year ago

Distinguish between zinc nitrate and lead nitrate

Kobotan [32]

Answer:

Zinc nitrate gives white ppt. which dissolves in excess ammonium hydroxide and produce a colorless solution whereas lead nitrate gives a chalky white ppt. of lead hydroxide which doesnot dissolve.

Explanation:

Hope this helps :)

7 0

2 years ago

Which one of the following statements about mixtures is correct [2] [3] The different substances in a mixture are always in a de

aivan3 [116]

Answer: Option (4) is the correct answer.

Explanation:

A mixture is defined as a substance that contains two or more different substance that are physically mixed with each other.

If solute particles are evenly distributed in a solvent then it is known as a homogeneous mixture.

For example, salt dissolved in water is a homogeneous mixture.

If solute particles are unevenly distributed into the solvent then it is known as a heterogeneous mixture.

For example, sand in water is a heterogeneous mixture.

Thus, we can conclude that the statement a mixture must contain at least two different substances, is correct about mixtures.

4 0

3 years ago

Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

4 0

3 years ago