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schepotkina [342]
2 years ago
5

Read "Embankment dam at a gold mine." Of the two opinions to consider in this case, which one presents a stronger argument? How

does that help you make your final decision as the consulting dam engineer? What was your decision for this dam, as the consultant? How can you accomplish this and still stay within your budget? (Site 1)
Chemistry
1 answer:
riadik2000 [5.3K]2 years ago
8 0

The mayor presents a stronger argument because he is concerned over the safety of the citizens. If the gold mine isn't as productive as it used to be, yet is still being used, then the dam should be repaired or replaced in some manner. For the safety of the village & for the continuation of the gold mine, my decision would be to repair the dam. To keep the budget cost low, I would only repair what is needed to be fixed.

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Which of the following shows a Bronsted-Lowry acid reacting?
7nadin3 [17]
<h3>Answer:</h3>

         Option-C:  HCl + H₂O  →   H₃O⁺ + Cl⁻

Explanation:

       Bronsted-Lowery concept of Acid and Base defines Acid as that specie which tends to donate H⁺ (Hydrogen Ion) and bases are those species which accepts H⁺ from Acids.

In selected option, HCl is reacting as Acid as it donates H⁺ to water (lowery bronsted base).

Also, the correspong acid is converted into conjugate base (i.e. Cl⁻) and base is converted into conjugate acid (i.e. H₃O⁺)

8 0
2 years ago
Read 2 more answers
From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2
sashaice [31]

Answer:

From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2(g)ΔH∘=−127.2kJ

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol and

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

(Ans)

ΔHf° of CaC2 = -59.0 kJ/mol

Explanation:

CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2 (g) = −127.2kJ

ΔHrxn = −127.2kJ

ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - ΔHf°(CaC2)- 2ΔHf°(H2O);

ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn

Where

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol and

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol

ΔHf°(CaC2) = -59.0 kJ/mol

7 0
2 years ago
What volume (ml) of a 0.2450 m koh(aq) solution is required to completely neutralize 55.25 ml of a 0.5440 m h3po4(aq) solution?
Nonamiya [84]
<span>Answer: It depends on what came after "0.5440 M H...". If it was a monoprotic acid, like HCl, the calculation would go like this: (55.25 mL) x (0.5440 M acid) x (1 mol KOH / 1 mol acid) / (0.2450 M KOH) = 122.7 mL KOH If it was a diprotic acid, like H2SO4, like this: (55.25 mL) x (0.5440 M acid) x (2 mol KOH / 1 mol acid) / (0.2450 M KOH) = 245.4 mL KOH If it was a triprotic acid, like H3PO4, like this: (55.25 mL) x (0.5440 M acid) x (3 mol KOH / 1 mol acid) / (0.2450 M KOH) = 368.0 mL KOH</span>
5 0
2 years ago
Under which conditions will sugar most likely dissolve fastest in a cup of water? A. a sugar cube with stirring at 5oC B. a suga
OverLord2011 [107]
I believe the answer is sugar crystals with stirring at 15 degrees Celsius. 
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5 0
2 years ago
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How many grams are in 2.5 pound sample
julsineya [31]
About 2,500 grams Ans balkfdoaks; 
5 0
2 years ago
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