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The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵
<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>
It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.
Hence, the equilibrium constant of the reaction in discuss is;
K = [5.6]²/[0.10]³[0.10]
k = 5.6² × 10⁴
k = 3.136 × 10⁵
K = 3.1 × 10⁵.
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the amount of heat produced from the combustion of 24.3 g benzene (c6h6) is ΔH = -976.5 kJ
There are two moles of benzene involved in the process (C6H6). Since the heat of this reaction is -6278 kJ, the burning of 2 moles of benzene will result in a heat loss of 6278 kJ. This reaction is exothermic.
Enthalpy, or the value of H, is a unit of measurement for heat that relies on the amount of matter present (number of moles).
Thus, 24.3 g of benzene contains:
n = mass/molar mass, where n = 24.3/78.11, and n = 0.311 moles.
2 moles = 6278 kJ
0.311 moles =x
By the straightforward direct three rule:
2x = -1953.08 x = -976.5 kJ
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Answer:
During a total lunar eclipse, the Earth lies directly between the sun and the moon, causing the Earth to cast its shadow on the moon.
Explanation:
Considering the Charles's law, the sample of carbon dioxide gas will occupy 308.72 mL.
<h3>Charles's law</h3>
Charles's law establishes the relationship between the temperature and the volume of a gas when the pressure is constant. This law says that the volume is directly proportional to the temperature of the gas: for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases.
Mathematically, Charles's law states that the ratio between volume and temperature will always have the same value:
Considering an initial state 1 and a final state 2, it is fulfilled:
<h3>Final volume in this case</h3>
In this case, you know:
- V1= 250 mL
- T1= 25 C= 298 K (being 0 C=273 K)
- V2= ?
- T2= 95 C= 368 K
Replacing in Charles's law:
Solving:
<u><em>V2= 308.72 mL</em></u>
Finally, the sample of carbon dioxide gas will occupy 308.72 mL.
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