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Aleksandr [31]
2 years ago
14

The compound Xe(CF3)2 decomposes in a first-order reaction to elemental Xe with a half-life of 30.0 min. If you place 4.5 mg of

Xe(CF3)2 in a flask, calculate how long you must wait until only 0.25 mg of Xe(CF3)2 remains?
Chemistry
1 answer:
GenaCL600 [577]2 years ago
7 0

Answer : The time passed by the sample is, 1.2\times 10^2\text{ min}

Explanation :

Half-life = 30.0 min

First we have to calculate the rate constant, we use the formula :

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

k=\frac{0.693}{30.0\text{ min}}

k=0.0231\text{ min}^{-1}

Now we have to calculate the time passed.

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

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

where,

k = rate constant  = 0.0231\text{ min}^{-1}

t = time passed by the sample  = ?

a = initial amount of the reactant  = 4.5 mg

a - x = amount left after decay process =0.25 mg

Now put all the given values in above equation, we get

t=\frac{2.303}{0.0231}\log\frac{4.5}{0.25}

t=125.15\text{ min}=1.2\times 10^2\text{ min}

Therefore, the time passed by the sample is, 1.2\times 10^2\text{ min}

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A flask contains methane, chlorine and carbon monoxide gases. the partial pressures of each are 0.215 atm, 0.066 atm, and 0.826
hjlf

Answer;

The total pressure is 1.107 atm.

Explanation;

The total pressure is the sum of the pressures of the three gases in the flask

Pressure (total) = 0.215 atm + 0.066 atm + 0.826 atm = 1.107 atm

= 1.107 atm.

8 0
2 years ago
Read 2 more answers
Which of the following statements concerning hydrocarbons is/are correct?
Alona [7]

Answer:

1.  Saturated hydrocarbons may be cyclic or acyclic molecules.

2.  An unsaturated hydrocarbon molecule contains at least one double bond.

Explanation:

Hello,

In this case, hydrocarbons are defined as the simplest organic compounds containing both carbon and hydrogen only, for that reason we can immediately discard the third statement as ethylenediamine is classified as an amine (organic chain containing NH groups).

Next, as saturated hydrocarbons only show single carbon-to-carbon bonds and carbon-to-hydrogen bonds, they may be cyclic (ring-like-shaped) or acyclic (not forming rings), so first statement is true

Finally, since we can find saturated hydrocarbons which have single carbon-to-carbon and carbon-to-hydrogen bonds only and unsaturated hydrocarbons which could have double or triple bonds between carbons and carbon-to-hydrogen bonds, the presence of at least one double bond makes the hydrocarbon unsaturated.

Therefore, first and second statements are correct.

Best regards.

6 0
2 years ago
A 100 mL reaction vessel initially contains 2.60×10^-2 moles of NO and 1.30×10^-2 moles of H2. At equilibrium the concentration
Sliva [168]

Answer:

<h2>The equilibrium constant Kc for this reaction is 19.4760</h2>

Explanation:

The volume of vessel used= 100 ml

Initial moles of NO= \frac{2.60}{10^2} moles

Initial moles of H2= \frac{1.30}{10^2} moles

Concentration of NO at equilibrium= 0.161M

Concentration(in M)=\frac{moles}{volume(in litre)}

Moles of NO at equilibrium= 0.161(\frac{100}{1000})

                                            =\frac{1.61}{10^2} moles

               

                    2H2 (g)        +    2NO(g) <—>    2H2O (g) +    N2 (g)

<u>Initial</u>          :1.3*10^-2          2.6*10^-2                0                   0        moles

<u>Equilibrium</u>:1.3*10^-2 - x     2.6*10^-2-x              x                   x/2     moles

∴\frac{2.60}{10^2}-x=\frac{1.61}{10^2}

⇒x=\frac{0.99}{10^2}

Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)

<u>Equilibrium</u>:0.31*10^-2      1.61*10^-2          0.99*10^-2        0.495*10^-2  moles

⇒Kc=\frac{(0.0099)^2(0.00495)}{(0.0031)^2(0.0161)^2}  (0.1)

⇒Kc=19.4760

3 0
2 years ago
Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.
Lesechka [4]

<u>Answer:</u> The enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

For the given chemical reaction:

C(s)+2H_2(g)\rightarrow CH_4(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]

We are given:

\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ

Hence, the enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

3 0
2 years ago
The Lewis structures of four compounds are given. In the first molecule sufur is the central atom. There are two oxygen atoms bo
vesna_86 [32]

Answer:

Compound 1. Sulfur Dioxide

Compound 3. Dichloromethane

Compound 4. Phosphorus Trichloride

<em>(figure attached)</em>

Explanation:

Compound 1. Sulfur Dioxide

Polar compound

Through the Lewis structure it is confirmed that SO₂ is a polar compound, because it is an asymmetric compound having two regions of different polarity. The lower region having oxygen groups is more electronegative then the upper region.

Compound 2. Carbon Dioxide

Non polar Compound

Through the Lewis structure it is confirmed that CO₂ is a non polar compound, because it is a symmetric compound having two regions of same polarity. The left region and the right region both contains oxygen groups having same electronegativity.

Compound 3. Dichloromethane

Polar compound

Through the Lewis structure it is confirmed that CH₂Cl₂ is a polar compound, because it is an asymmetric compound having two regions of different polarity. Two chlorine atoms are attached to it and as we know that chlorine is a more electronegative element than hydrogen so it attracts the bonding pair of electrons towards itself which creates polarity.

Compound 4. Phosphorus Trichloride

Through the Lewis structure it is confirmed that PCl₃ is a polar compound, because three chlorine atoms attached to it and as we know that chlorine is a more electronegative element so it attracts the bonding pair of electrons towards itself which creates polarity.

8 0
2 years ago
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