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svet-max [94.6K]
3 years ago
8

Converting forestland to urban areas through urban development could lead to which of the following? A. Less land available for

lumber production. B. Reduced demand for trees. C. Reduced lumber prices. D. Less costly ways of growing trees.
Chemistry
2 answers:
ahrayia [7]3 years ago
8 0

Answer: Option (a) is the correct answer.

Explanation:

Conversion of forestland into urban areas requires the cutting of large number of trees so that more land could be available for the urban area.

This cutting of trees will result into reduced demand for trees. Therefore, there will be less land available for lumber production.

Thus, we can conclude that option (a) is the correct answer.

geniusboy [140]3 years ago
3 0
The answer for APEX is A. Less land available for lumber production 
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Which statement concerning the hydrated hydrogen ion is incorrect? (a) Although we often represent hydrogen ions as bare protons
Harrizon [31]

Answer:C. The value of n for H+(H2O)n can be calculated for almost all solutions.

Explanation:

An hydrate can be described as a substance that contains water or with an hydrogen bonded water molecule group.

The hydrate group doesn't necessarily have a fixed formula.

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3 years ago
Construct a three-step synthesis of trans-2-pentene from acetylene by dragging the appropriate formulas into the bins. Note that
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Answer:

The three-step synthesis of trans-2-pentene from acetylene is as follows.

<u>Step -1:</u> Formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkanes.

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<u>Step -3:</u> Formation of trans-pent - 2-pent-ene by reduction.

Explanation:

Synthesis of trans-pent-2-yne from ethyne takes place is mainly a three step synthesis which involves formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkane. Second step involves the further alkylation of terminal alkynes to higher order nonterminal alkynes and the third step involves the formation of trans-2-ene by dissolving reduction method.

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8 0
3 years ago
When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t
OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

Mn(s)+2HCl(aq)\rightarrow  MnCl_2(aq)+H_2(g)

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

m=1.00 g/mL\times 100.0 mL = 100 g

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

q=m\times c\times (T_{final}-T_{initial})

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = 4.18 J/^oC

T_{final} = final temperature = 23.1^oC

T_{initial} = initial temperature = 28.9^oC

Now put all the given values in the above formula, we get:

q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC

q=2,242.4 J=2.242 kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = \frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol

\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

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