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lozanna [386]
3 years ago
11

For the reaction 2Fe + 3Cl2 → 2FeCl3 which of the following options gives the correct reactant:reactant ratio? Fe:Cl2 = 1:1 Fe:C

l2 = 2:3 Fe:Cl2 = 3:2 Fe:FeCl3 = 1:1
Chemistry
1 answer:
BigorU [14]3 years ago
8 0

Answer: Fe:Cl2 = 2:3

Explanation:

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Write the general chemical formula for an acid, and discuss what each component of the formula represents.
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The general formula of an acid is HX where H+ acts as the cation while X can be a halogen (Cl-, Fl-, Br-, etc) and acts as the anion. There are also cases where X is an ion like SO42-, which brings two atoms of H in the formula. Strong acids dissociate completely while weak acids do not.
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What is the sequence of the complementary strand for the DNA ,gcaa
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The density of a gas is 1.25 grams/liter. Which of the following is correct to convert the density of the gas to ounce/millilite
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How do I do this question?The aluminum cup inside your calorimeter weighs 39.78 g. You add 50.01 g of ice cold water to the calo
konstantin123 [22]

Answer:

Cp_{metal}=0.922\frac{J}{g\°C}

Explanation:

Hello.

In this problem we must realize that there is heat flow that moves from the hot metal object and the hot water to the cold water and the cold aluminum cup, which allows us to write:

Q_{cup}+Q_{cold,w}=-(Q_{metal}+Q_{hot,w})

Which means that the heat lost be the hot metal object and the hot water is gained by both the cold water and the cold aluminum cup, which can be written in terms of mass, specific heats and change in temperature towards the equilibrium temperature (35.9 °C):

m_{cup}Cp_{cup}(T_{eq}-T_{cup})+m_{cold,w}Cp_{cold,w}(T_{eq}-T_{cold,w})=-(m_{metal}Cp_{metal}(T_{eq}-T_{metal})+m_{hot,w}Cp_{hot,w}(T_{eq}-T_{hot,w})

We need to solve for the specific heat of the metal as shown below:

Cp_{metal}=\frac{m_{cup}Cp_{cup}(T_{eq}-T_{cup})+m_{cold,w}Cp_{cold,w}(T_{eq}-T_{cold,w})+m_{hot,w}Cp_{hot,w}(T_{eq}-T_{hot,w})}{-m_{metal}(T_{eq}-T_{metal})} \\\\Cp_{metal}=\frac{39.78g*0.903\frac{J}{g\°C}(35.9-0.5)\°C+50.01g*4.184\frac{J}{g\°C}(35.9-0.5)\°C +50.72g*4.184\frac{J}{g\°C}(35.9-69.5)\°C  }{-49.98g(35.9-69.5)\°C } \\\\Cp_{metal}=\frac{1271.6J+7407.2J-7130.3J}{-1679.3g\°C} \\\\Cp_{metal}=0.922\frac{J}{g\°C}Best regards.

3 0
3 years ago
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