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2 years ago

The combustion of ethyne, shown below unbalance, produces heat which can be used to weld metals:

Chemistry

1 answer:

Andreyy892 years ago

3 0

Answer:

3.69 g

Explanation:

Given that:

The mass m = 325 g

The change in temperature ΔT = ( 1540 - 165)° C

= 1375 ° C

Heat capacity $c_p$ = 0.490 J/g°C

The amount of heat required:

q = mcΔT

q = 325 × 0.490 × 1375

q = 218968.75 J

q = 218.97 kJ

The equation for the reaction is expressed as:

$C_2H_{2(g)} + 5O_{2(g)} \to 2CO_{2(g)} + H_2O_{(g)} \ \ \ \ \ \Delta H^o_{reaction} = -1544 \ kJ$

Then,

1 mole of the ethyne is equal to 26 g of ethyne required for 1544 kJ heat.

Thus, for 218.97 kJ, the amount of ethyne gas required will be:

$= \dfrac{26 \ g}{1544 \ kJ} \times 218.97 \ kJ$

= 3.69 g

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) Based on the graph, determine the order of the decomposition reaction of cyclobutane at 1270 K. Justify your answer.

lidiya [134]

Answer:

- The order of the reaction based on the graph provided is first order.

- 99% of the cyclobutane would have decomposed in 53.15 milliseconds.

d) Rate = K [Cl₂]

K = rate constant

The justification is presented in the Explanation provided below.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Explanation:

To investigate the order of a reaction, a method of trial and error is usually employed as the general equations for the amount of reactant left for various orders are known.

So, the behaviour of the plot of maybe the concentration of reactant with time, or the plot of the natural logarithm of the concentration of reactant with time.

The graph given is evidently an exponential function. It is a graph of the concentration of cyclobutane declining exponentially with time. This aligns with the gemeral expression of the concentration of reactants for a first order reaction.

C(t) = C₀ e⁻ᵏᵗ

where C(t) = concentration of the reactant at any time

C₀ = Initial concentration of cyclobutane = 1.60 mol/L

k = rate constant

The rate constant for a first order reaction is given

k = (In 2)/T

where T = half life of the reaction. It is the time taken for the concentration of the reactant to fall to half of its initial concentration.

From the graph, when the concentration of reactant reaches half of its initial concentration, that is, when C(t) = 0.80 mol/L, time = 8.0 milliseconds = 0.008 s

k = (In 2)/0.008 = (0.693/0.008) = 86.64 /s

Calculate the time, in milliseconds, that it would take for 99 percent of the original cyclobutane at 1270 K to decompose

C(t) = C₀ e⁻ᵏᵗ

when 99% of the cyclobutane has decomposed, there only 1% left

C(t) = 0.01C₀

k = 86.64 /s

t = ?

0.01C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.01

In e⁻ᵏᵗ = In 0.01 = -4.605

-kt = -4.605

t = (4.605/k) = (4.605/86.64) = 0.05315 s = 53.15 milliseconds.

d) The reaction mechanism for the reaction of cyclopentane and chlorine gas is given as

Cl₂ → 2Cl (slow)

Cl + C₅H₁₀ → HCl + C₅H₉ (fast)

C₅H₉ + Cl → C₅H₉Cl (fast)

The rate law for a reaction is obtained from the slow step amongst the the elementary reactions or reaction mechanism for the reaction. After writing the rate law from the slow step, any intermediates that appear in the rate law is then substituted for, using the other reaction steps.

For This reaction, the slow step is the first elementary reaction where Chlorine gas dissociates into 2 Chlorine atoms. Hence, the rate law is

Rate = K [Cl₂]

K = rate constant

Since, no intermediates appear in this rate law, no further simplification is necessary.

The obtained rate law indicates that the reaction is first order with respect to the concentration of the Chlorine gas and zero order with respect to cyclopentane.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Hope this Helps!!!

8 0

1 year ago

What is the formal charge on the nitrogen in hydroxylamine, h2noh?

andrew11 [14]

<h3>Answer:</h3>

Formal Charge on Nitrogen is "Zero".

<h3>Explanation:</h3>

Formal Charge on an atom in molecules is calculated using following formula;

Formal Charge = [# of Valence e⁻s] - [e⁻s in lone pairs + 1/2 # of Bonding e⁻s]

As shown in attached picture of Hydroxylamine, Nitrogen atom is containing two electrons in one lone pair of electrons and six electrons in three single bonds with two hydrogen and one oxygen atom respectively.

Hence,

Formal Charge = [5] - [2 + 6/2]

Formal Charge = [5] - [2 + 3]

Formal Charge = 5 - 5

Formal Charge = 0 (zero)

Hence, the formal charge on nitrogen atom in hydroxylamine is zero.

5 0

1 year ago

2CH4(g)⟶C2H4(g)+2H2(g)

Rasek [7]

Answer : The enthalpy change for the reaction is, 201.9 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The balanced reaction of $CH_4$ will be,

$2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)$ $\Delta H^o=?$

The intermediate balanced chemical reaction will be,

(1) $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-890.3kJ$

(2) $C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)$ $\Delta H_2=-136.3kJ$

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=-571.6kJ$

(4) $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)$ $\Delta H_4=-3120.8kJ$

Now we will multiply the reaction 1 by 2, revere the reaction 2, reverse and half the reaction 3 and 4 then adding all the equations, we get :

(1) $2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)$ $\Delta H_1=2\times (-890.3kJ)=-1780.6kJ$

(2) $C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)$ $\Delta H_2=-(-136.3kJ)=136.3kJ$

(3) $H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)$ $\Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ$

(4) $2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)$ $\Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ$

The expression for enthalpy of the reaction will be,

$\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)$

$\Delta H=201.9kJ$

Therefore, the enthalpy change for the reaction is, 201.9 kJ

5 0

1 year ago

Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b

frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ ?

(a) $Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}$

(b) $2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}$

(d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Answer: (d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Explanation: Redox Reaction is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called cell notation, formed by: left side of the salt bridge (||), which is always the anode, i.e., its half-equation is as an oxidation and right side, which is always the cathode, i.e., its half-equation is always a reduction.

For the cell notation: $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$

Aluminum's half-equation is oxidation:

$Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}$

For Lead, half-equation is reduction:

$Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}$

Multiply first half-equation for 2 and second half-equation by 3:

$2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}$

$3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}$

Adding them:

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

The balanced redox reaction with cell notation $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ is

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

6 0

2 years ago

Heating 235 g of water from 22.6°C to 94.4°C in a microwave oven requires 7.06 × 104 J of energy. If the microwave frequency is

Darya [45]

Answer: 3.69 × 10^27

Explanation:

Amount of energy required = 7.06 × 10^4 J

Frequency of microwave (f) = 2.88 × 10^10 s−1

Planck's constant (h) = 6.63 × 10^-34 Jᐧs/quantum

Recall ;

Energy of photon = hf

Therefore, energy of photon :

(6.63 × 10^-34)j.s× (2.88 × 10^10)s^-1

= 19.0944 × 10^(-34 + 10) = 19.0944×10^-24 J

Hence, number of quanta required :

(7.06 × 10^4)J / (19.0944 × 10^-24)J

= 0.369 × 10^(4 + 24) = 0.369×10^28

= 3.69 × 10^27

6 0

2 years ago