Answer: 238.6 J
Explanation:
1) Chemical energy is indicated as enthalpy
2) Energy balance:
∑ enthalpy of the reactants + energy added = ∑ enthalpy of the products + energy released.
3) ∑ enthalpy of the reactants = 85.1 J + 87.9 J = 173 J
4) energy added = 104.3 J
5) ∑ enthalpy of the products = 38.7 J + D
6) energy released = 0
7) Equation:
173J + 104.3J = 38.7 + D + 0
⇒ D = 173J + 104.3J - 38.7J = 238.6J, which is the chemical energy of the product D.
Answer:
m = 7.39 g.
Explanation:
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In this case, since the molar mass of iron (III) phosphate is 150.82 g/mol based on its molecular formula (FePO₄), we can compute the mass in grams of 0.0490 moles of this compound by setting up the following dimensional analysis:
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You must use the Periodic Table. Metals, on the right side, will be positive. Non-metals, left, are negative. You need to know the specific ionic charge elements.
Answer:
b. hydride shift from C-3 to C-2.
Explanation:
Markovnikov's rule states that *in the addition of a protic acid HX or other polar reagent to an asymmetric alkene, the acid hydrogen (H) or electropositive part gets attached to the carbon with more hydrogen substituents, and the halide (X) group or electronegative part gets attached to the carbon with more alkyl substituents* (wikipedia).
This rule implies that the hydrogen of HBr will be attached to C-1 and the carbocation will be on C-2. Remember that the order of stability of carbocations is tertiary > secondary > primary > methyl. A hydride shift can yield a tertiary carbocation.
C-3 is a tertiary carbon atom. If the hydride on carbon 3 shifts to carbon 2, a tertiary and more stable carbocation is formed. This accounts for the major product in the reaction.
Answer is: distillation.
Solution of sugar and water has a uniform composition and is able to be separated by physical means.
Solution is homogeneous mixture has the same proporties of its components throughout any given sample.
Solution of water and salt, can be separated with distillation (process of heating and cooling). Water evaporates and sugar remain in the beaker.