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ArbitrLikvidat [17]
1 year ago
5

A liquid with high viscosity _____ flow easily and _____ effective in wetting a surface.

Chemistry
1 answer:
scoundrel [369]1 year ago
5 0

A liquid with high  viscosity does not flow easily and is not effective in wetting a surface.

When a metal is subjected to corrosive elements including salt, moisture, and high temperatures, a reaction called corrosion takes place inside the metal. Some foods contain metallic compounds that can corrode a material. The majority of corrosion is simply surface dis-colouration, which polishing agents may quickly remove.

Increasing viscosity and constant intermolecular water bonding together result in surface tension. Any liquid that was more viscous than water possessed a surface tension that was equal to or lower than that of water. Viscosity with surface tension decrease when temperature rises.

Therefore, a liquid with high viscosity does not flow easily and is not effective in wetting a surface.

To know more about viscosity

brainly.com/question/2193315

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Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili
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<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}

\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M

\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

                         H_2+I_2\rightleftharpoons 2HI

<u>Initial:</u>                0.1    0.1

<u>At eqllm:</u>          0.1-x   0.1-x   2x

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\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M

The expression of K_c for above equation follows:

K_c=\frac{[HI]^2}{[H_2][I_2]}

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[I_2]_{eq}=0.0210M

Putting values in above expression, we get:

K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61

Hence, the value of equilibrium constant for the given reaction is 56.61

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