Answer:
It is just slightly less abundant than its alkali cousin, sodium. Potassium is less dense than water, so it can float on water. However, chemically, potassium reacts with water violently. It will give off hydrogen and eventually catch fire.
Let the ratio of grams of hydrogen per gram of carbon in methane be M, we know that:
M = 0.3357 g / 1 g
Next, lets represent the grams of hydrogen per gram of carbon in ethane be E. The final piece of information we have is:
M / E = 4/3
If we cross multiply,
3M = 4E
Now, substituting the value of M from earlier and solving for E,
E = (3 * 0.3357) / 4
E = 0.2518
There are 0.2518 grams of hydrogen per gram of carbon in ethane.
Answer: 0.52849 j /g °C
Explanation:
Given the following :
Mass of metal = 36g
Δ Temperature of metal = (28.4 - 99)°C = - 70.6°C
Mass of water = 70g
Δ in temperature of water = (28.4 - 24.0) = 4.4°C
Heat lost by metal = (heat gained by water + heat gained by calorimeter)
Quantity of heat(q) = mcΔT
Where; m = mass of object ; c = specific heat capacity of object
Heat lost by metal:
- (36 × c × - 70.6) = 2541.6c - - - - (1)
Heta gained by water and calorimeter :
(70 × 4.184 × 4.4) + (12.4 × 4.4) = 1288.672 + 54.56 = 1343.232 - - - - (2)
Equating (1) and (2)
2541.6c = 1343.232
c = 1343.232 / 2541.6
c = 0.52849 j /g °C
Answer:
The correct answer is 32.2 grams.
Explanation:
Based on the given information, the enthalpy of formation for aluminum oxide is 1676 kJ/mol. It signifies towards the energy that is required to generate aluminum and oxygen, and both of these exhibit zero enthalpy of formation. Therefore, the ΔHreaction is the required energy to generate 2 moles of aluminum. Thus, the energy needed for the formation of single mole of aluminum is,
ΔHrxn = 1676/2 = 838 kJ/mol
Q or the energy input mentioned in the given case is 1000 kJ. Therefore, the number of moles of Al generated is,
(1000 kJ) / (838 kJ/Al mole) = 1.19 moles of Aluminum
The grams of aluminum produced can be obtained by using the formula,
mass = moles * molecular mass
= 1.19 * 26.98
= 32.2 grams.
Answer:
Explanation:
2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s); ΔᵣH = ?
The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is
2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s)
ΔfH°/kJ·mol⁻¹: 0 -824.3 -1675.7 0
![\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [1(-1675.7) + 2(0)] - [2(0) - 1(-824.3)]\\& = & -1675.7 + 824.3\\& = & \textbf{-851.4 kJ/mol}\\\end{array}\\\text{The enthalpy change is } \large \boxed{\textbf{-851.4 kJ/mol}}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5CDelta_%7B%5Ctext%7Br%7D%7DH%5E%7B%5Ccirc%7D%20%26%20%3D%20%26%20%5B1%28-1675.7%29%20%2B%202%280%29%5D%20-%20%5B2%280%29%20-%201%28-824.3%29%5D%5C%5C%26%20%3D%20%26%20-1675.7%20%2B%20824.3%5C%5C%26%20%3D%20%26%20%5Ctextbf%7B-851.4%20kJ%2Fmol%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20enthalpy%20change%20is%20%7D%20%5Clarge%20%5Cboxed%7B%5Ctextbf%7B-851.4%20kJ%2Fmol%7D%7D)
