Answer: -227 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%20n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BC_2H_2%7D%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]](https://tex.z-dn.net/?f=-1255.8%3D%5B%282%5Ctimes%20-393.5%29%2B%281%5Ctimes%20-241.8%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%28%5Cfrac%7B5%7D%7B2%7D%5Ctimes%200%29%5D)
![-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]](https://tex.z-dn.net/?f=-1255.8%3D%5B%28-787%29%2B%28-241.8%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%280%29%5D)
Therefore, the enthalpy change for 
is -227 kJ.
Answer:
8
Explanation:
To solve this question, we just need to put the new number into the equation. If [A] remain constant then that mean [A2]= [A1]. If B doubled, then that mean [B2]= 2[B2]. To find what factor does the rate of reaction increases, we need to divide the first reaction rate with the second. The calculation will be:
rate2/rate1= k[A2][B2]³ / k[A1][B1]³
rate2/rate1= [A1][2B1]³ / [A1][B1]³
rate2/rate1= A1*8B1³ / A1*B1³
rate2/rate1= 8/1= 8
The rate of reaction will be 8 times faster.
The answer is 2 mol of H₂O will be produced.
The balanced equation for the chemical reaction is:
<span>c3h8 (g) + 5o2 (g) → 3co2 (g) + 4h2o (g)
5 moles of O</span>₂ produces 4 moles of H₂O, and when there is 2.5 mol of O₂, moles of H₂O will be:
2.5 x 4/5 = 2 mol of H₂O
Answer:
D
Explanation:
The noble gases are said to be inert because they undergo very few chemical reactions. They enjoy their status by having completely filled electron shells and do not have any reason whatsoever to go into chemical reactions. Most elements otherwise go into chemical reactions principally to enjoy the stability of these elements.
Noble gases can be found in the last group of the periodic table.
Answer:
The final volume is 39.5 L = 0.0395 m³
Explanation:
Step 1: Data given
Initial temperature = 200 °C = 473 K
Volume = 0.0250 m³ = 25 L
Pressure = 1.50 *10^6 Pa
The pressure reduce to 0.950 *10^6 Pa
The temperature stays constant at 200 °C
Step 2: Calculate the volume
P1*V1 = P2*V2
⇒with P1 = the initial pressure = 1.50 * 10^6 Pa
⇒with V1 = the initial volume = 25 L
⇒with P2 = the final pressure = 0.950 * 10^6 Pa
⇒with V2 = the final volume = TO BE DETERMINED
1.50 *10^6 Pa * 25 L = 0.950 *10^6 Pa * V2
V2 = (1.50*10^6 Pa * 25 L) / 0.950 *10^6 Pa)
V2 = 39.5 L = 0.0395 m³
The final volume is 39.5 L = 0.0395 m³
