1- metal and non metal
2- true
3- chlorine
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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The total amount of heat required is the sum of all the sensible heat and latent heats involved in bringing the ice to a desired temperature and state. The latent heat of fusion and vaporization of water 333.55 J/g and 2260 J/g, respectively. Solving for the total amount of heat,
total amount of heat = 13.0 g (2.09 J/gC)(12) + 13(333.55 J/g) + 13.0 g (4.18 J/gC)(100 - 0) + (13.0 g)(2260 J/g) + (13 g)(2.01 J/g)(113-100)
= 39815.88 J
= 39.82 kJ
Answer:
grams H₂O produced = 8.7 grams
Explanation:
Given 2C₂H₆(g) + 7O₂(g) => 4CO₂(g) + 6H₂O(l)
7g 18g ?g
Plan => Convert gms to moles => determine Limiting reactant => solve for moles water => convert moles water to grams water
Moles Reactants
moles C₂H₆ = 7g/30g/mol = 0.233mol
moles O₂ = 18g/32g/mol = 0.563mol
Limiting Reactant => (Test for Limiting Reactant) Divide mole value by respective coefficient of balanced equation; the smaller number is the limiting reactant.
moles C₂H₆/2 = 0.233/2 = 0.12
moles O₂/7 = 0.08
<u><em>Limiting Reactant is O₂</em></u>
Moles and Grams of H₂O:
Use Limiting Reactant moles (not division value) to calculate moles of H₂O.
moles H₂O = 6/7(moles O₂) = 6/7(0.562) moles H₂O = 0.482 mole H₂O yield
grams H₂O = (0.482mol)(18g·mol⁻¹) = 8.7 grams H₂O
Answer:
Chemical change.
Explanation:
When it is heated it decomposes into mercury and oxygen gas. The mercury oxide reactant becomes the silver color of mercury. Hence, a color change can be noticed throughout the reaction.
