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Korvikt [17]
4 years ago
13

The given graduated cylinder is calibrated in

Chemistry
2 answers:
Marina CMI [18]4 years ago
8 0

Answer : The volume of liquid in graduated cylinder is 26.5 mL.

Explanation :

As we know that for the measurement of the volume of liquid in graduated cylinder are shown by placing the graduated cylinder on the flat surface and then view the height of the liquid in the graduated cylinder with the naked eyes directly level with the liquid.

The liquid will tend to curve downward that means this curve is known as the meniscus.

In the case of colored liquid, we are always read the upper meniscus of the liquid for the measurement.

In the case of colorless liquid, we are always read the lower meniscus of the liquid for the measurement.

In the given image, there are 10 division between the 20 and 30 and the solution is colorless. So, we will read the lower meniscus of the liquid for the measurement.

Hence, the volume of liquid in graduated cylinder is 26.5 mL.

matrenka [14]4 years ago
5 0

Answer: 26mL

Explanation:

You always need to see the meniscus of the liquid, (which is the curved figure in the liquid) in the cylinder, also, the question is telling you that write the correct number of significant figures, as the cylinder does not point at any decimal number, you cannot tell if it is on 26.1 or 26.9, you can be certain, only, that is 26 mL.

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describe the number of protons, neutrons, and electrons in the atom, where each type of particle is located​
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Answer:

Protons - positive charge located in the nucleus

Neutrons - neutral charge located in the nucleus

Electrons - negative charge located outside the nucleus (they orbit)

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3 years ago
How are a mold and cast related?
White raven [17]
They are both related to hold the shape or a part of an organism.
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3 years ago
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What is the mass, in grams, of one mole of any substance known as?
Elena-2011 [213]

Answer: molar mass

Explanation:

The molar mass of any substance is it's relative molecular mass expressed in grams. Hence, the relationship between molar mass of any substance and mass, in grams, of one mole of any substance is mathematically expressed as:

Number of moles =

(Mass in grams ➗ Molar mass)

Hence, the unit of molar mass is gram per mole (g/mol)

Thus, molar mass is the answer.

3 0
3 years ago
In an experiment, 2.54 grams of copper completely reacts with sulfur, producing 3.18 grams of copper(I) sulfide.
kozerog [31]

Answer:

               0.64 g of S

Solution:

               The balance chemical equation is as follow,

                                           2 Cu + S ----> Cu₂S

According to equation,

                        127 g (2 mole) Cu produces = 159 g (1 mole) of Cu₂S

So,

                                 2.54 g Cu will produce = X g of Cu₂S

Solving for X,

                     X = (2.54 g * 159 g) / 127 g

                     X = 3.18 g of Cu₂S

Now, it is confirmed that the reaction is 100% ideal. Therefore,

As,

                       127 g (2 mole) Cu required = 32 g (1 mole) of S

So,

                                2.54 g Cu will require = X g of S

Solving for X,

                      X = (2.54 g * 32 g) / 127 g

                      X = 0.64 g of S

5 0
3 years ago
Read 2 more answers
Propane has a normal boiling point of -42.04 °C and a heat of vaporization of 24.54 kJ/mole. What is the vapor pressure of propa
Mademuasel [1]

Answer : The vapor pressure of propane at 25.0^oC is 17.73 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of propane at 25.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 25.0^oC=273+25.0=298.0K

T_2 = normal boiling point of propane = -42.04^oC=230.96K

\Delta H_{vap} = heat of vaporization = 24.54 kJ/mole = 24540 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{24540J/mole}{8.314J/K.mole}\times (\frac{1}{298.0K}-\frac{1}{230.96K})

P_1=17.73atm

Hence, the vapor pressure of propane at 25.0^oC is 17.73 atm.

3 0
3 years ago
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