Answer: ΔH for the reaction is -277.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%28products%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%28reactant%29%5D)
![\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCCl_4%7D%5Ctimes%20%5CDelta%20H_%7BCCl_4%7D%29%2B%28n_%7BHCl%7D%5Ctimes%20B.E_%7BHCl%7D%29%20%5D-%5B%28n_%7BCH_4%7D%5Ctimes%20%5CDelta%20H_%7BCH_4%7D%29%2Bn_%7BCl_2%7D%5Ctimes%20%5CDelta%20H_%7BCl_2%7D%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20-139%29%2B%281%5Ctimes%20-92.31%29%20%5D-%5B%281%5Ctimes%20-74.87%29%2B%281%5Ctimes%20121.0%5D)
Therefore, the enthalpy change for this reaction is, -277.4 kJ
Answer:
9.1 mol
Explanation:
The balanced chemical equation of the reaction is:
CO (g) + 2H2 (g) → CH3OH (l)
According to the above balanced equation, 2 moles of hydrogen gas (H2) are needed to produce 1 mole of methanol (CH3OH).
To convert 36.7 g of hydrogen gas to moles, we use the formula;
mole = mass/molar mass
Molar mass of H2 = 2.02g/mol
mole = 36.7/2.02
mole = 18.17mol
This means that if;
2 moles of H2 reacts to produce 1 mole of CH3OH
18.17mol of H2 will react to produce;
18.17 × 1 / 2
= 18.17/2
= 9.085
Approximately to 1 d.p = 9.1 mol of methanol (CH3OH).
The molarity is moles/liters.
First, convert 4,000 mL to L:
4000 mL --> 4 L
Now, you must convert the 17 g of solute to moles by dividing the number of grams by the molar mass. The molar mass of AgNO3 is <span>169.87 g/mol:
17 / 169.87 = .1
Now that you have both the number of moles and the liters, plug them into the initial equation of moles/liters:
.1/4 = .025</span>
1. C
2. C
3. In elastic deformation, the deformed body returns to its original shape and size after the stresses are gone. In ductile deformation, there is a permanent change in the shape and size but no fracturing occurs. In brittle deformation, the body fractures after the strength is above the limit.
4. Normal faults are faults where the hanging wall moves in a downward force based on the footwall; they are formed from tensional stresses and the stretching of the crust. Reverse faults are the opposite and the hanging wall moves in an upward force based on the footwall; they are formed by compressional stresses and the contraction of the crust. Thrust faults are low-angle reverse faults where the hanging wall moves in an upward force based on the footwall; they are formed in the same way as reverse faults. Last, Strike-slip faults are faults where the movement is parallel to the crust of the fault; they are caused by an immense shear stress.
I hope this helped :D
Explanation:
The two half equations are;
3e + HNO3 → NO
S→ H2SO4 + 6e
When balancing half equations, we have to make sure the number of electrons gained is equal to the number of electrons lost.
<em>Which factor will you use for the top equation?</em>
We multiply by 2 to make the number of electrons = 6e
<em>Which factor will you use for the bottom equation?</em>
We multiply by 1 to make the number of electrons = 6e
