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meriva
3 years ago
13

Answer the following question using the reaction: AB ---> A+B

Chemistry
1 answer:
Vadim26 [7]3 years ago
3 0

Answer:

So whats the question?

Explanation:

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5. A sample of methane has a volume of 225 mL at 0.650 atm of pressure. What will the new volume be if the
Norma-Jean [14]

Answer:

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Explanation:

8 0
2 years ago
a chemist dissolved an 11.9-g sample of koh in 100.0 grams of water in a coffee cup calorimeter. when she did so, the water temp
Snowcat [4.5K]

The heat of solution is -51.8 kJ/mol

<h3>What is the heat of solution?</h3>

We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.

Number of moles of KOH =  11.9-g/56 g/mol = 0.21 moles

Temperature rise = 26.0 ∘c

Mass of the water = 100.0 grams

Heat capacity =  4.184 j/g⋅°c

Then;

ΔH = mcθ

ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ

Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol

Learn more about heat of solution:brainly.com/question/24243878

#SPJ1

4 0
2 years ago
Consider the reaction
SOVA2 [1]

Answer :

(a) The average rate will be:

\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s

(b) The average rate will be:

\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)

The expression for rate of reaction :

\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}

\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}

\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}

\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}

\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}

Thus, the rate of reaction will be:

\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}

<u>Part (a) :</u>

<u>Given:</u>

\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s

As,  

-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}

and,

\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}

\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s

\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s

<u>Part (b) :</u>

<u>Given:</u>

\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s

As,  

-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}

and,

-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}

\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s

\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s

5 0
3 years ago
2. Consider the reaction 2 Cg H18 (4) +250â (9) ⺠16 co, (g) + 18 HâO(g) la How many moles of H20co) are produced, when |--16:1
Sladkaya [172]

Answer :

(a) The moles of water produced are 145.35 moles.

(b) The mass of oxygen needed are 3080.8 grams.

<u>Solution for part (a) : Given,</u>

Moles of C_8H_{18} = 16.15 moles

First we have to calculate the moles of H_2O

The balanced chemical reaction is,

2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O

From the balanced reaction we conclude that

As, 2 moles of C_8H_{18} react to give 18 moles of H_2O

So, 16.15 moles of C_8H_{18} react to give \frac{16.15}{2}\times 18=145.35 moles of H_2O

The moles of water produced are 145.35 moles.

<u>Solution for part (b) : Given,</u>

Mass of C_8H_{18} = 878 g

Molar mass of C_8H_{18} = 114 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_8H_{18}.

\text{ Moles of }C_8H_{18}=\frac{\text{ Mass of }C_8H_{18}}{\text{ Molar mass of }C_8H_{18}}=\frac{878g}{114g/mole}=7.702moles

Now we have to calculate the moles of O_2

The balanced chemical reaction is,

2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O

From the balanced reaction we conclude that

As, 2 moles of C_8H_{18} react with 25 moles of O_2

So, 7.702 moles of C_8H_{18} react with \frac{7.702}{2}\times 25=96.275 moles of O_2

Now we have to calculate the mass of O_2.

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(96.275moles)\times (32g/mole)=3080.8g

The mass of oxygen needed are 3080.8 grams.

6 0
3 years ago
Sailin' Steve likes to test the buoyancy of different objects. He threw a spherical object into a fluid that has a greater densi
Hunter-Best [27]
The spherical object will float and the square object will sink to the bottom of the fluid as it has greater density. :)
7 0
3 years ago
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