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just olya [345]

3 years ago

9

What substance is 52.4 % potassium and 47.6 % chlorine?

1 answer:

dimaraw [331]3 years ago

4 0

Polymer sand is correct

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Recall all the models you described in task 1. Think about the results each model would predict for Thomson’s experiment. Which

Bumek [7]

Answer:

J.J. Thomson’s experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons. Thomson proposed the plum pudding model of the atom, which had negatively-charged electrons embedded within a positively-charged “soup.”

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3 years ago

PLEASE ANSWER QUESTION 6!!

kap26 [50]

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Ps: TYSON all can come for talking about

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Consider the reaction below. C2H4(g) + H2(g) to C2H6(g) Which change would likely cause the greatest increase in the rate of the

algol [13]

D. Increase temperature and increase pressure

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3 years ago

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On a potential energy diagram for the following processes, which of the following has an increase in entropy?

daser333 [38]

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I believe that your answer would be <span>B. water freezes
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A 1.800-g sample of solid phenol (C6H5OH(s)) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/?C. The temp

vichka [17]

Answer:

The balanced chemical equation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat of combustion per gram of phenol is 32.454 kJ/g

Heat of combustion per gram of phenol is 3,050 kJ/mol

Explanation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat capacity of calorimeter = C = 11.66 kJ/°C

Initial temperature of the calorimeter = $T_1= 21.36^oC$

Final temperature of the calorimeter = $T_2= 26.37^oC$

Heat absorbed by calorimeter = Q

$Q=C\times \Delta T$

Heat released during reaction = Q'

Q' = -Q ( law of conservation of energy)

Energy released on combustion of 1.800 grams of phenol = Q' = -(58.4166 kJ)

Heat of combustion per gram of phenol:

$\frac{Q'}{1.800 g}=\frac{-58.4166 kJ}{1.800 g}=32.454 kJ/g$

Molar mass of phenol = 94 g/mol

Heat of combustion per gram of phenol:

$\frac{Q'}{\frac{1.800 g}{94 g/mol}}=\frac{-58.4166 kJ\times 94 g/mol}{1.800 g}=3,050 kJ/mol$

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3 years ago