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QveST [7]
2 years ago
15

Which of the following is the best piece of laboratory glassware for preparing 500.0 mL of an aqueous solution of a solid?(A) Vo

lumetric flask(B) Erlenmeyer flask(C) Test tube(D) Graduated beaker(E) Graduated cylinder
Chemistry
2 answers:
zubka84 [21]2 years ago
7 0

Answer:

Volumetric flask

Explanation:

When preparing a standard solution, a volumetric flask is the best piece of equipment required. The volumetric flask is a special equipment designed with a mark. The mark shows the required volume of solution.

The solute is added to the flask and simply made up to mark with distilled water to prepare a solution of the required volume in a particular volumetric flask of that volume.

professor190 [17]2 years ago
3 0

Answer: Option A. Volumetric flask

Explanation: volumetric flask is used in preparing standard solutions to specific or definite.

Note: Graduated beaker can only be used to estimate roughly the volume of a liquid.

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The combustion of any hydrocarbon yields water and carbon dioxide. We will now construct a balanced equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Each mole of propane requires 5 moles of oxygen.
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Why can metallic elements bend, while solid but non-metallic break or crumble?
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Metallic elements can bend because they have the property of being ductile. This means that a solid material stretches under tensile stress. If a material is ductile then it may be stretched into a wire. Further, the material can also be malleable. 
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What other objects could be used to simulate radioactive and nonradioactive nuclei? Check all that apply.
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a computer that shows pictures of atoms on screen

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2 years ago
A sample of n2 gas occupies 2.40 l at 20°c. if the gas is in a container that can contract or expand at constant pressure, at wh
Bas_tet [7]
From Charle's law the volume of a fixed mass of a gas is directly proportional to the absolute temperature at constant pressure.
Therefore';
V1/T1=V2/T2
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A compound is made up of 28 g N, 24 g C, 48 g O, and 8 g H .What is the empirical formula?
vovikov84 [41]

Answer:

\rm C_2H_8N_2O_3.

Explanation:

<h3>Step One: calculate the coefficients. </h3>

Look up the relative atomic mass of these four elements on a modern periodic table:

  • \rm C: approximately 12.
  • \rm H: approximately 1.
  • \rm N: approximately 14.
  • \rm O: approximately 16.

The relative atomic mass of an element is numerically equal to the mass (in grams, \rm g,) of one mole of atoms of this element.

For example, the relative atomic mass of \rm C is approximately 12. Therefore, each mole of \rm C\! atoms would have a mass of 12\; \rm g.

This sample contains 24\; \rm g of carbon. That would correspond to approximately \displaystyle \left(\frac{24}{12}\right)\; \rm mol = 2\; \rm mol of \rm C atoms.

Similarly, for the other three elements:

\displaystyle n(\mathrm{H}) \approx \frac{8\; \rm g}{1\; \rm g \cdot mol^{-1}} = 8\; \rm mol.

\displaystyle n(\mathrm{N}) \approx \frac{28\; \rm g}{14\; \rm g \cdot mol^{-1}} = 2\; \rm mol.

\displaystyle n(\mathrm{O}) \approx \frac{48\; \rm g}{16\; \rm g \cdot mol^{-1}} = 3\; \rm mol.

Hence, the ratio between these elements in this compound would be:

n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3.

In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.

n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3 is indeed the smallest possible integer ratio between the number of atoms of these elements.

<h3>Step Two: arrange the elements in an appropriate order</h3>

Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:

If carbon, \rm C, is present in this compound, then:

  • \rm C (carbon) and then \rm H (hydrogen) will be the first two elements listed in the formula (ignore the hydrogen if it is not in the compound.)
  • The other elements in this compound will be listed in alphabetical order.

If there is no carbon \rm C in this compound, then list all the elements in this compound in alphabetical order.

Both \rm C (carbon) and \rm H (hydrogen) are found in this compound. Therefore, the first element in the list would be \rm C\!. The second would be \rm H\!, followed by \rm N\! and then \rm O\!.

Hence, the empirical formula of this compound would be \rm C_2H_8N_2O_3.

6 0
2 years ago
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