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Sphinxa [80]
3 years ago
7

If a piece of marble has the mass of 132.796 g and the volume of 26.9 cm

Chemistry
2 answers:
Mashutka [201]3 years ago
8 0

Answer:

4.94g/cm3.

Explanation:

4.94g/cm3.

jek_recluse [69]3 years ago
6 0

Answer:4.93665 g/cm“3“

Explanation: 132.796g times 26.9 equals

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A scientist is conducting a Sanger's sequencing experiment to determine the number of polypeptides present in an oligomeric prot
seropon [69]

Answer:

Depends on molecule.

Explanation:

The number of the polypeptide chains present in the oligomer depends on the molecule. Some molecules have more polypeptide chains whereas some of them have less polypeptide chains. For example, Hemoglobin is a oligomer that consists of four Polypeptide Chains, two of these Polypeptide Chains are α-globin molecules, each comprise of 141 amino acids, and the other two are (β, γ, δ, or ε) globins, each consist of 146 amino acids.

7 0
3 years ago
Describe how particles of iron (Fe) an oxygen (O2) react to how produce oxide (Fe2O3), also known as rust.
forsale [732]

Answer:

Described by a redox reaction below

Explanation:

Iron(III) oxide is an ionic compound, since it consists of a metal, iron, and a nonmetal, oxygen.

Ionic compounds are formed when metals lose their valence electrons in order to have an octet in their previous shell and donate them to nonmetal atoms, so that nonmetals fill their outer shell to have an octet.

As a result, positive ions (cations) and negative ions (anions) are formed. When iron reacts with oxygen, the following reaction takes place:

4 Fe (s) + 3 O_2 (g)\rightarrow 2 Fe_2O_3 (s)

This is a redox (oxidation–reduction) reaction, since we have electron loss and gain. Four iron atoms lose a total of 12 electrons to obtain a +3 charge in the final compound, while 3 oxygen molecules gain these 12 electrons to become 6 oxide anions with a -2 charge.

4 0
3 years ago
If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution
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This is a straightforward dilution calculation that can be done using the equation

M_1V_1=M_2V_2

where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

M_2=\frac{M_1V_1}{V_2}.

Substituting in our values, we get

\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

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Is it the dry lab/wet lab week 1 or ?
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