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2 years ago

Which molecule can't be exist?A)NCl3 B)OF6 C)ONF3 D)PCl3

Chemistry

1 answer:

Usimov [2.4K]2 years ago

4 0

Answer:

OF6

Explanation:

Oxygen and fluorine have almost the same electronegativity, so oxygen isn't able to donate enough electron density to fluorine to form OF6. Atomic size. There isn't enough room around an oxygen atom for six fluorine atoms.

:-befrank

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What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 4

geniusboy [140]

Answer:

5.195 ml

Explanation:

Dissociation of HCl yields;

HCl(aq) ⇄ H⁺(aq) + Cl⁻(aq)

∴ [HCl] = [H⁺] = [Cl⁻]

pH = 1.9

= -log[ H⁺]

= $10^{-pH}$

= $10^{-1.9}$

= 0.01258 M

Moles of HCl = number of mole × volume

= 0.01258 × 4.80 L

= 0.06042 mole

Mass of HCl = number of moles of HCl × Molar mass of HCl

Mass of HCl = 0.06042 mol × 36.5 g/mol

Mass of HCl = 2.205 grams

Also, given that the density = 1.179 g/mL

Mass of HCl in the solution = 36% of the density

i.e $\frac{36}{100}*1.179$

= 0.4244 grams

Volume of a concentrated HCl solution needed can now be determined as;

= $\frac{2.205}{0.4244}$

= $5.195 ml$

7 0

3 years ago

Iridium has only two naturally occurring isotopes. The mass of iridium-191 is 190.9605 amu and the mass of iridium-193 is 192.96

cluponka [151]

Answer:

Abundance of Iridium-193 is 62.75%

Explanation:

From the question given above, the following data were obtained:

Isotope A (Iridium-191):

Mass of A = 190.9605 amu

Abundance of A = A%

Isotope B (Iridium-193):

Mass of B = 192.9629 amu

Abundance B = (100 – A) %

Relative atomic mass of Iridium = 192.217 amu

Next, we shall determine the abundance of isotope A (Iridium-191). This can be obtained as follow:

Relative atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]

192.217 = [(190.9605 × A%)/100] + [(192.9629 × (100 – A)%)/100]

192.217 = 1.909605A% + 1.929629(100 – A)%

192.217 = 1.909605A% + 192.9629 – 1.929629A%

Collect like terms

192.217 – 192.9629 = 1.909605A% – 1.929629A%

–0.7459 = –0.020024A%

Divide both side by –0.020024

A% = –0.7459 / –0.020024

A% = 37.25 %

Finally, we shall determine the abundance of Isotope B (Iridium-193).

This can be obtained as follow:

Abundance of A (Iridium-191) = 37.25 %

Abundance of B (Iridium-193) =?

Abundance B = 100 – A%

Abundance B = 100 – 37.25 %

Abundance of B (Iridium-193) = 62.75%

4 0

3 years ago

Filtration, devastation, and evaporation are examples of

Mariana [72]

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3 years ago

Carbon dioxide in the atmosphere can be reduced by _.*

pychu [463]

Answer:

As the use of plants as carbon sinks can be undone by events such as wildfires, the long-term reliability of these approaches has been questioned. Carbon dioxide that has been removed from the atmosphere can also be stored in the Earth's crust by injecting it into the subsurface, or in the form of insoluble carbonate salts (mineral sequestration).

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Explanation:

6 0

3 years ago

Read 2 more answers

Metal cations are atoms that have a positive charge. What type of bond is a metal cation likely to form? A. covalent bond B. dou

Lunna [17]

Answer:

D. ionic bond

Explanation:

Due to electron deficiency in a metal cation, they cannot form a covalent bond beacuse it means to share electrons. By contrast metal cation seek for electrons. In an ionic bond, one atom give electrons, while another atom recevie electron. Because of that, this is the better option to metal cations.

4 0

3 years ago

Which molecule can't be exist?A)NCl3 B)OF6 C)ONF3 D)PCl3​

Which molecule can't be exist?A)NCl3 B)OF6 C)ONF3 D)PCl3