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Harlamova29_29 [7]
3 years ago
7

1. As you increase speed is there an increase or decrease in Kinetic Energy?

Chemistry
1 answer:
drek231 [11]3 years ago
3 0

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

1. As you increase speed is there an increase or decrease in Kinetic Energy?

\qquad \sf  \dashrightarrow \: increase

2. As you increase mass is there an increase or decrease in Kinetic Energy?

<h3 />

\qquad \sf  \dashrightarrow \: increase

3. As you decrease speed is there an increase or decrease in Kinetic Energy?

\qquad \sf  \dashrightarrow \: decrease

4. As you decrease mass is there an increase or decrease in Kinetic Energy?

\qquad \sf  \dashrightarrow \: decrease

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Mg + _LINO3 → _Mg(NO3)2 + _Li
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What is the theoretical yield of this test reaction? Record answer to the nearest whole number.
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Answer : The theoretical yield of water in this reaction is 360 g.

Solution : Given,

Mass of

Mass of

Mass of

Molar mass of  = 44 g/mole

Molar mass of  = 24 g/mole

Molar mass of  = 18 g/mole

First we have to calculate the moles of  and .

Moles of  =

Moles of  =

The given balanced chemical reaction is,

From the given reaction, we conclude that

1 mole of  react with 2 mole of

20 moles of  requires  of

But the actual moles of  = 42 moles

Excess moles of  = 42 - 40 = 2 moles

So,  is the limiting reagent.

Now we have to calculate the mass of .

From the reaction, we conclude that

1 mole of  gives 1 mole of

20 moles of  gives 20 moles of

Mass of  = Moles of  × Molar mass of  

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Read more on Brainly.com - brainly.com/question/2525928#readmore

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Ayuda Ejercicio de FISICOQUIMICA<br><br> Ayuda con procedimiento y solucion <br><br> Bendiciones
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sry but I don't know chemistry but I can help you in other subjects ,can I do any other favour for u.

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3 years ago
A 28.4 g sample of an unknown metal is heated to 39.4 °C, then is placed in a calorimeter containing 50.0 g of water. Temperatur
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Answer:

THE SPECIFIC HEAT OF THE METAL IS 0.8983 J/g °C

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In solving the problem, we have to understand that:

Heat lost by the metal = Heat gained by the water in the bomb calorimeter

First is to calculate the heat evolved from the reaction

Heat = mass * specific heat * change in temperature

Mass of water = 50 g

specific heat of water = 4.184 J/g °C

Change in temperature = 23 - 21 = 2 °C

So therefore,

Heat = 50 * 4.184 * 2

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Next is to solve for the specific heat of the metal;

Heat lost by the metal is the same as the heat gained by water

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Change in temperature = 39.4 °C - 23 °C = 16.4 °C

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C = 418.4 / 465.76

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The specific heat of the metal is hence 0.8983 J/g °C

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