CHEMISTRY COLLEGE

In a constantvolume bomb calorimeter, the combustion of 0.6654 gof an organic
compound with a molecular mass of 46.07 amu causesthe temperature
in the calorimeter to rise from 25.000oC to 30.589
oC. The total heat capacity ofthe calorimeter and all
its contents is 3576 JoC-1. What is
the energy of combustion ofthe organic compound,
DU/ kJ
mol-1?

Answers

Answer 1

Answer:

Answer:

1383.34 kJ/mol is the energy released on combustion of the organic compound.

Explanation:

Mass of an organic compound = 0.6654 g

Molar mass of organic compound = 46.07 g/mol

Moles of an organic compound = $(0.6654 g)/(46.07 g/mol)=0.01444 mol$

Let heat evolved during burning of 0.6654 grams of an organic compound be -Q.

Heat absorbed by calorimeter = Q' = -Q

The total heat capacity of the calorimeter all its contents = C

C = 3576 J/°C

Change in temperature of the calorimeter =

ΔT = 30.589°C - 25.000°C = 5.589°C

$Q'=C* \Delta T$

$Q'=3576 J/^oC* 5.589^oC=19,975.536 J=19.975 kJ$

Q' = 19.975 kJ

Q = -19.975 kJ (negative sign; energy released)

0.01444 moles of an organic compound gives 19.975 kilo Joule.

The 1 mole of an organic compound will give : $\Delta H_(comb)$

$\Delta H_(comb)=(-19.975 kilo Joule)/(0.01444 mol)$

$=-1383.34 kJ/mol$

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Calculate the molar solubility of mercury (I) bromide, Hg2Br2, in 1.0 M KBr. The Ksp for Hg2Br2 is 5.6 X 10−23. (Hint: How would the Br− concentration from the sparingly soluble compound itself compare to the Br− concentration that comes from the KBr?

Answers

Answer:

The correct answer is 5.6 × 10⁻²³ M.

Explanation:

As a highly soluble salt, KBr dissolves easily in water, while Hg₂Br₂ is very less soluble in comparison to KBr.

Let the solubility of Hg₂Br₂ is S mol per liter.

Therefore,

KBr (s) (1.0 M) ⇒ K⁺ (aq) (1M) + Br⁻ (aq) (1M)

Hg₂Br₂ (s) (1-S) ⇔ Hg₂⁺ (aq) (S) + 2Br⁻ (aq) (2S)

Net [Br-] = (2S + 1) M

Ksp = S (2S + 1)²

Ksp = S (4S² + 1 + 4S)

Ksp = 4S³ + S + 4S²

As the solubility is extremely less, therefore, we can ignore S² and S³. Now,

Ksp = S = 5.6 × 10⁻²³ M

Hence, the solubility of Hg₂Br₂ is 5.6 × 10⁻²³ M.

How many significant digits are in 89015?

Answers

Answer:5

Explanation:

What is the mass of silver chlorate (191.32 g/mol) that decomposes to release 0.466L of oxygen gas at STP? AgC1036) _AgCl). _026) A) 1.33g B) 597 E) 7.968 C) 3.988 D) 2658

Answers

Answer : The mass of silver chlorate will be 2.654 grams.

Explanation :

The balanced chemical reaction is,

$2AgClO_3\rightarrow 2AgCl+3O_2$

First we have to calculate the moles of oxygen gas at STP.

As, 22.4 L volume of oxygen gas present in 1 mole of oxygen gas

So, 0.466 L volume of oxygen gas present in $(0.466)/(22.4)=0.0208$ mole of oxygen gas

Now we have to calculate the moles of silver chlorate.

From the balanced chemical reaction, we conclude that

As, 3 moles of oxygen produced from 2 moles of silver chlorate

So, 0.0208 moles of oxygen produced from $(2)/(3)* 0.0208=0.01387$ moles of silver chlorate

Now we have to calculate the mass of silver chlorate.

$\text{Mass of }AgClO_3=\text{Moles of }AgClO_3* \text{Molar mass of }AgClO_3$

Molar mass of silver chlorate = 191.32 g/mole

$\text{Mass of }AgClO_3=0.01387mole* 191.32g/mole=2.654g$

Therefore, the mass of silver chlorate will be 2.654 grams.

A balloon filled with 0.500 L of air at sea level is submerged in the water to a depth that produces a pressure of 3.25 atm. What is the volume of the balloon at this depth? a. 0.154 L b. 6.50 L c. 0.615 L d. 1.63 L d. None of the above

Answers

"0.154 L" is the volume of the balloon.

Given:

Pressure,

$P_1 = 1 \ atm$
$P_2 = 3.25 \ atm$

Volume,

$V_1 = 0.5 \ L$
$V_2 = ?$

As we know,

→ $P_1. V_1 = P_2 .V_2$

or,

→ $V_2 = (P_1. V_1)/(P_1)$

By substituting the values, we get

$= (0.5* 1)/(3.25)$

$= (0.5)/(3.25)$

$= 0.154 \ L$

Thus the above answer i.e., "option a" is correct.

Learn more:

brainly.com/question/16365001

Answer:

Option a . 0.154L

Explanation:

P₁ . V₁ = P₂ . V₂

when we have constant temperature and constant moles for a certain gas.

At sea level, pressure is 1 atm so:

0.5 L . 1atm = V₂ . 3.25 atm

(0.5L . 1atm) / 3.25 atm = 0.154 L

The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (M) Initial Rate (10–3 M/s) 0.4 0.4 0.2 160 0.2 0.4 0.4 80 0.6 0.1 0.2 15 0.2 0.1 0.2 5 0.2 0.2 0.4 20 Using the initial-rate method, what is the order of the reaction with respect to C? a. zero-order b. first-order c. third-order d. second-order e. impossible to tell from the data given

Answers

The dependence of the power of the reaction rate on the concentration is called the order of the reaction. The order of the reaction is the first order.

What is the initial-rate method?

The initial rate method is the estimation of the order of the reaction by the initial rates of the reactants and products and by performing the reaction several times by measuring the rate.

The reaction is given as,

$\rm A + B + C \rightarrow Products$

The rate of reaction can be given as:

$\rm rate = k[A]^(x)[B]^(y)[C]^(z)$

Here the variables x, y and z are orders respective to the reactant concentration and k is the rate constant.

Value of x with respect to A:

$\begin{aligned} \rm \frac {Rate 3}{Rate 4} &= \rm [([A(3)])/([A(4)])]^(\rm x)\n\n(15)/(5) &= [([0.6])/([0.2])]^(\rm x)\n\n\rm x &= 1\end{aligned}$

Value of y with respect to B:

$\begin{aligned}\rm \frac {Rate 2}{Rate 5} &= \rm [([B(2)])/([B(5)])]^(\rm y)\n\n(80)/(20) &= [([0.4])/([0.2])]^(\rm y)\n\n\rm y &= 2\end{aligned}$

Value of z with respect to C:

$\rm \frac {Rate 1}{Rate 2} &= [([A(1)])/([A(2)])]^(x) [([B(1)])/([B(2)])]^(y) [([C(1)])/([C(2)])]^(z)$

Substituting value of x = 1 and y = 2 in the above equation:

$\begin{aligned}(160)/(80) &= [([0.4])/([0.2])]^(1)[([0.4])/([0.4])]^(2) [([0.2])/([0.4])]^(\rm z)\n\n1 &= (0.5)^(\rm z)\n\n&= 1\end{aligned}$

Therefore option b. with respect to C = 1, the order of the reaction is first-order.

Learn more about the order of reaction here:

brainly.com/question/8139015

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

[A]₀ [B]₀ [C]₀ Initial Rate (10⁻³ M/s)

1. 0.4 0.4 0.2 160

2. 0.2 0.4 0.4 80

3. 0.6 0.1 0.2 15

4. 0.2 0.1 0.2 5

5. 0.2 0.2 0.4 20

The rate of the above reaction is given as:

$Rate = k[A]^(x)[B]^(y)[C]^(z)$

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

$(Rate3)/(Rate4)= [([A(3)])/([A(4)])]^(x)$

$(15)/(5)= [([0.6])/([0.2])]^(x)$

$3 = 3^(x) \n\nx =1$

Order w.r.t B : Use trials 2 and 5

$(Rate2)/(Rate5)= [([B(2)])/([B(5)])]^(y)$

$(80)/(20)= [([0.4])/([0.2])]^(y)$

$4 = 2^(y) \n\ny =2$

Order w.r.t C : Use trials 1 and 2

$(Rate1)/(Rate2)= [([A(1)])/([A(2)])]^(x)[([B(1)])/([B(2)])]^(y)[([C(1)])/([C(2)])]^(z)$

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

$(160)/(80)= [([0.4])/([0.2])]^(1)[([0.4])/([0.4])]^(2)[([0.2])/([0.4])]^(z)$

$1 = (0.5)^(z)$

z = 1

Therefore, order w.r.t C = 1

In the reaction C zH4 + H2 - e) +4 CHo the carbon atoms are a) oxidized b) reduced c) cannot be determined

Answers

Answer: Option (b) is the correct answer.

Explanation:

Reduction is defined as the process in which there occurs gain of hydrogen. Whereas oxidation is defined as the process in which there occurs loss of hydrogen.

As the given reaction is as follows.

$C_(2)H_(4) + H_(2) \rightarrow C_(2)H_(6)$

Since, hydrogen is being added in this chemical reaction. It means that reduction is taking place and carbon atom is reduced.

Thus, we can conclude that in the given reaction carbon atoms are reduced.

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