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3 years ago

What is the rate law?

Chemistry

1 answer:

nikdorinn [45]3 years ago

7 0

Answer: B

Explanation:

The rate law is the mathematical equation that describes how reactant concentration changes as a function of time. A law such as "Rate = k*[A]*[B]" means that, for each liter-equivalent of the reactant(s) A, there are k liters of reactant B. The law also dictates the molarity (and thus partial pressure) for each component in solution.

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Part A Write balanced molecular equation for the reaction between nitric acid and calcium hydroxide. Express your answer as a ch

Basile [38]

Answer:

2HNO3+Ca(OH)2 = Ca(NO3)2+2H2O

Explanation:

The reaction between Nitric acid(HNO3)and Calcium hydroxide(Ca(OH)2) gives Calcium Nitrate( Ca(NO3)2 and Water( H2O)

7 0

3 years ago

A 5.00-cm cube of magnesium has a mass of 217.501 g. What is the density of magnesiummetal?

Doss [256]

Answer:

d = 43.5 g/cm³

Explanation:

Given data:

Mass of magnesium cube = 217.501 g

Volume of magnesium cube = 5.00 cm³

Density of magnesium cube = ?

Solution:

Formula:

d = m/v

d = density

m = mass

v = volume

by putting values,

d = 217.501 g/ 5.00 cm³

d = 43.5 g/cm³

8 0

3 years ago

How is science used on a daily basis give 3 reasons how

goldenfox [79]

1-Autonomy: Our desire to direct our own lives. In short: “You probably want to do something interesting, let me get out of your way!”
2-Mastery: Our urge to get better at stuff.
3-Purpose: The feeling and intention that we can make a difference in the world.

7 0

3 years ago

The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

Part 1 :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Part 2 :

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Part 3 :

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Part 4 :

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

3 years ago

what is the density of mercury if 205 mL has a mass of 2790 g; round your answer to the nearest 0.1 g/mL

kherson [118]

D = m / V

D = 2790 g / 205 mL

D = 13.60 g/mL

4 0

3 years ago