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NNADVOKAT [17]
4 years ago
9

Indicate the direction of polarity of each of the covalent bonds by placing the appropriate delta notation next to each end of t

he bond. C-O, O-CL, O-F, C-N, C-L, S-H, S-CL
Chemistry
1 answer:
o-na [289]4 years ago
6 0

Answer:

C→O, O→Cl, O→F, C→N, C←Li, S ←H, S→Cl

Explanation:

The general convention for indicating polarity in molecules is that of showing the direction of polarity from the positive to the negative part of the bond. This is indicated in the answer by arrows that show the direction of the dipole.

The negative end of the dipole must be the more electronegative element while the positive end of the dipole is the less electronegative element, hence the answer shown above.

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What is the reason why there are so many organic compounds?
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3 years ago
An 100g box have a temperature of 22 C. What temperature must a 200g box be kept at in order for both to have the same amount of
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3 years ago
An unknown mass of aluminum requires 6290 joules to raise
Alex Ar [27]

Answer:

124.579 g

Explanation:

Amount of heat required to change the temperature of body is given by the equation given below

Q = m* c * ΔT ____________________equation A

where Q is the total heat energy required by the any object

c is the specific heat capacity of the body J/gK

ΔT is the difference of final temperature of the object and initial temperature of the object

______________________________________

Given Q = 6290 joules

c = 0.900 J/gK

To calculate the temperature in kelvin from Celsius

we can use the formula

K = C + 273

initial temperature  = 23.9°C

initial temperature on kelvin scale =  23.9° + 273 = 296.9

Final temperature  = 80°C

initial temperature on kelvin scale =  80° + 273 = 353

ΔT(temperature difference) = 353 - 296.9 = 56.1

___________________________________________

let the mass of of the  aluminum be m g

substituting the value of Q , c, ΔT in equation A we have

Q = m* c * ΔT

=> 6290 = m * 0.900 * 56.1

=> m = 6290/(0.900 * 56.1 )

=> m = 6290/50.49

=> m = 124.579 g

The mass of the  aluminum is 124.579 g

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