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4 years ago

9

The decomposition of kclo3 proceeds spontaneously when it is heated. do you think that the reverse reaction, the formation of kc

lo3 from kcl and o2, is likely to be feasible under ordinary conditions? explain your answer.

1 answer:

pishuonlain [190]4 years ago

7 0

<span>The answer is No, it is not possible that the reverse reaction will occur at ordinary conditions. This is because the entropy of the KCl and the O2 are much largely negative than that of the KClO3. The entropy component of the Gibbs free energy assures that the Gibbs free energy for the reverse reaction is positive (Positive Gibbs means nonspontaneous reaction). </span>

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why is whenever one raised to the power of any exponent 1 and why is 0 raised to the power of any exponent 1

IceJOKER [234]

When you use exponents, think of it like this. 1 squared is 1 x 1. 1 cubed is 1 x 1 x 1. And 1 to the power of 4 is 1 x 1 x 1 x 1. And so on. You basically just multiply them by themselves. 0 to the power of any exponent is 1, well that's just a rule. It doesn't make much sense but it's easy to remember and I wouldn't worry about it.

5 0

3 years ago

If we start with 1.000 g of strontium-90, 0.805 g will remain after 9.00 yr. This means that the of strontium-90 is ________ yr.

lesantik [10]

The question is incomplete, here is the complete question.

If we start with 1.000 g of strontium-90, 0.805 g will remain after 9.00 yr. This means that the half-life of strontium-90 is ________ yr.

a. 28.8 b. 30.9 c. 35.4 d. 32.2

Answer : The half-life of strontium-90 is 28.8 years.

Explanation :

This is a type of radioactive decay and all radioactive decays follow first order kinetics.

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant

t = time taken for decay process = 9.00 year

a = initial amount or moles of the reactant = 1.000 g

a - x = amount or moles left after decay process = 0.805 g

Putting values in above equation, we get:

$k=\frac{2.303}{9.00}\log\frac{1.00}{0.805}$

$k=0.0241\text{ year}^{-1}$

To calculate the half-life, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$0.0241\text{ year}^{-1}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=28.8\text{ years}$

Therefore, the half-life of strontium-90 is 28.8 years.

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3 years ago

Rank the following substances in order from most soluble in water to least soluble in water: methane, CH4; 2-butanol, C4H9OH; ma

inna [77]

Answer:

MgCl2 > C4H9OH > CH4 > C3H8.

Explanation:

Alkanes do not form hydrogen bonds and are insoluble in polar solvents e.g water. The hydrogen bonds between water molecules are move away from an alkane molecule and this worsens as their Carbon chain / molecular weight increases.

MgCl2 is soluble in water. Water essentially breaks down the ionic crystal lattice and the resulting solution is slightly basic.

Alcohols are generally soluble in water and this is because of the -OH group and its ability to form hydrogen bonds with water molecules. As applied to alkanes, as the carbon chain in the alkyl group increases, the solubility decreases.

From the most soluble to the least soluble,

MgCl2 > C4H9OH > CH4 > C3H8.

4 0

3 years ago

Consider the equation:

Dmitry [639]

Answer:1. $Rate=k[CHCl_3]^1[Cl_2]^\frac{1}{2}$

2. The rate constant (k) for the reaction is $3.50M^\frac{-1}{2}s^{-1}$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$rate=k[CHCl_3]^x[Cl_2]^y$

k= rate constant

x = order with respect to $CHCl_3$

y = order with respect to $Cl_2$

n = x+y= Total order

1. a) From trial 1: $0.0035=k[0.010]^x[0.010]^y$ (1)

From trial 2: $0.0069=k[0.020]^x[0.010]^y$ (2)

Dividing 2 by 1 : $\frac{0.0069}{0.035}=\frac{k[0.020]^x[0.010]^y}{k[0.010]^x[0.010]^y}$

$2=2^x,2^1=2^x$ therefore x=1.

b) From trial 2: $0.0069=k[0.020]^x[0.010]^y$ (3)

From trial 3: $0.0098=k[0.020]^x[0.020]^y$ (4)

Dividing 4 by 3: $\frac{0.0098}{0.0069}=\frac{k[0.020]^x[0.020]^y}{k[0.020]^x[0.010]^y}$

$1.4=2^y,2^{\frac{1}{2}}=2^y$ therefore $y=\frac{1}{2}$

$rate=k[CHCl_3]^1[Cl_2]^\frac{1}{2}$

2. to find rate constant using trial 1:

$0.0035=k[0.010]^1[0.010]^\frac{1}{2}$

$k=3.50M^\frac{-1}{2}s^{-1}$

5 0

4 years ago

Why we use ethanol and salt in the titration ?

frutty [35]

Cause its the strongest force inside water

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3 years ago