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sdas [7]
3 years ago
11

¿Por qué los gases nobles tienen tendencia a no formar compuesto si como todo elemento de la tabla periódica estos tienen electr

ones de valencias los cuales garantizan los enlaces.?
Chemistry
1 answer:
White raven [17]3 years ago
6 0
Noble gases are the least reactive of all elements. That's because they have eight valence electrons, which fill their outer energy level. This is the most stable arrangement of electrons, so noble gases rarely react with other elements and form compounds.
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Iron sulfite has the formula?
Mademuasel [1]
Its FeSO3 or iron(iii)sulfite = Fe2(SO3)3
7 0
3 years ago
Calculate the hight of a column of liquid glycerol in meters required to exert the same pressure as 3.02 m if CCl4?
katovenus [111]

Answer:

Approximately 3.81\; \rm m.

Explanation:

Look up the density \rho of carbon tetrachloride, \rm CCl_4, and glycerol:

  • Density of carbon tetrachloride: approximately 1.59\times 10^{3}\; \rm kg \cdot m^{-3}.
  • Density of glycerol: approximately 1.26\times 10^{3}\; \rm kg \cdot m^{-3}.

Let g denote the gravitational field strength. (Typically g \approx 9.81\; \rm N \cdot kg^{-1} near the surface of the earth.) For a column of liquid with a height of h, if the density of the liquid is \rho, the pressure at the bottom of the column would be:

P = \rho\cdot g \cdot h.

The pressure at the bottom of this carbon tetrachloride column would be:

\begin{aligned} P &= \rho \cdot g \cdot h \\ & \approx 1.59\times 10^{3} \; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1} \times 3.02 \; \rm m \approx 4.71 \times 10^{4} \; \rm N \cdot m^{-2} \end{aligned}.

Rearrange the equation P = \rho\cdot g \cdot h for h:

\displaystyle h = \frac{P}{\rho \cdot g}.

Apply this equation to calculate the height of the liquid glycerol column:

\begin{aligned}h &= \frac{P}{\rho \cdot g} \\ &\approx \frac{4.71 \times 10^{4}\; \rm N \cdot m^{-2}}{1.26 \times 10^{3}\; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1}} \approx 3.81\; \rm m\end{aligned}.

4 0
2 years ago
In lab (write this down in your lab protocol), you will be given a stock solution that has a glucose concentration of 60 mg/dL.
Wittaler [7]

Answer:

1. The dilution factor for the serial dilution = 2

2. V2 = 1 mL

3. V1 = 0.5 mL

Explanation:

1. Dilution factor is the ratio of the initial concentration to the final concentration.

Dilution factor = initial concentration / final concentration

First dilution: initial concentration = 60 mg/dL

final concentration = 30 mg/dL

Dilution factor = 60 mg/dL / 30 mg/dL = 2

Second dilution: initial concentration = 30 mg/dL

final concentration = 15 mg/dL

Dilution factor = 30 mg/dL / 15 mg/dL = 2

Therefore, the dilution factor for the serial dilution = 2

2. From the dilution formula, C1V1 = C2V2; V2 = final volume to be prepared.

Since 1 mL of the various glucose solutions are to be prepared, the final concentration, V2 = 1 mL

3. From the dilution formula, C1V1 = C2V2; V1 = initial concentration of the solution to be prepared.

C1/C2 = V2/V1

Since the dilution factor, C1/C2 is 2, V2/V1 = 2

V1 = V2/2

V1 = 1 mL / 2

V1 = 0.5 mL

6 0
3 years ago
The compound ammonium sulfate consists of two ions, NH4+ and SO42–, both of which are
julsineya [31]
They are both (polyatomic) ions.
6 0
2 years ago
Helpppppppppppppppppppppppppppp
arlik [135]

Answer:

can you send link

Explanation:

8 0
1 year ago
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