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3 years ago

99 points

Chemistry

2 answers:

svlad2 [7]3 years ago

7 0

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ It would be a covalent bond most likely. It cannot be an ionic bond as the two elements have the same number of electrons. It also couldn't be a metallic bond as they are obviously not metals.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

frez [133]3 years ago

3 0

The following is true for electronegativity and the bonds formed between atoms:

If the difference in electronegativity is between 0 to 0.4, then the type of bond formed will likely be covalent.

If the difference is between 0.5 to 1.7, the bond is likely going to be polar covalent.

If the difference is greater than 1.8, then the bond will likely be ionic.

The difference in electromagnetically is 1.0 (4.0-3.0=1.0). Thus, the bond will be polar covalent, which is a type of covalent bond.

Let me know if you need any clarifications, thanks!

~ Padoru

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Considering the ideal gas law, the volume of gas produced at 25.0 °C and 1.50 atm is 184.899 L.

<h3>Definition of ideal gas</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

<h3>Ideal gas law</h3>

An ideal gas is characterized by absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of gases:

P×V = n×R×T

<h3>Volume of gas</h3>

In this case, you know:

P= 1.50 atm
V= ?
n= 500 g× $\frac{1 mole}{44 g}$ = 11.36 moles, being 44 $\frac{g}{mole}$ the molar mass of CO₂
R= 0.082 $\frac{atmL}{molK}$
T= 25 C= 298 K (being 0 C=273 K)

Replacing in the ideal gas law:

1.50 atm×V = 11.36 moles×0.082 $\frac{atmL}{molK}$ × 298 K

Solving:

V= (11.36 moles×0.082 $\frac{atmL}{molK}$ × 298 K) ÷ 1.50 atm

Finally, the volume of gas produced at 25.0 °C and 1.50 atm is 184.899 L.

Learn more about the ideal gas law:

<u>brainly.com/question/4147359?referrer=searchResults</u>

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