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3 years ago

Why couldn't you substitute 3m H2SO4 for concentrated HNO3 when oxidizing copper?

2 answers:

5 0

This question seems to be an essay question from experiment. Different solution of oxidizing agent will have different strength. Sulfuric acid or H2SO4 is weaker oxidizing agent when compared to nitric acid (HNO3). In this case, if you subtitute the H2SO4 you wouldn't be able to get the same result for the experiment.

V125BC [204]3 years ago

5 0

<h2>Answer:</h2>

Every oxidizing agent has different properties and they are specific for different reagents. Different solution of oxidizing agent will have different strength.

Sulfuric acid or H2SO4 is weaker oxidizing agent when compared to nitric acid (HNO3). so when you use H2SO4 for the oxidizing copper it would not work because copper need strong oxidizing agent.

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Why is the temperature of products likely to be lower than the temperature of reactants in an endothermic reaction?

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2 years ago

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3 years ago

What is the mass in grams of 394 mL of ethylene glycol? Express your answer with the appropriate units.

AleksandrR [38]

Answer : The mass of ethylene glycol is, 437.34 grams

Solution : Given,

Volume of ethylene glycol = 394 ml

Density of ethylene glycol = 1.11 g/ml (Standard value)

Formula used :

$Density=\frac{Mass}{Volume}$

Now put all the given values in this formula, we get the mass of ethylene glycol.

$1.11g/ml=\frac{Mass}{394ml}$

$Mass=437.34g$

Therefore, the mass of ethylene glycol is, 437.34 grams

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3 years ago

What is the main component of atoms that determines the type of bond that forms between atoms or compounds? Explain.

goldenfox [79]

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3 years ago

The value of ΔG°′ΔG°′ for the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) is +1.67 kJ/mol+1.67 kJ/mol . If t

dedylja [7]

Answer:

The concentration of fructose-6-phosphate F6P ≅ 1.35 mM

Explanation:

Given that:

ΔG°′ is the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) = +1.67 kJ/mol = 1670 J/mol

concentration of glucose-6-phosphate at equilibrium = 2.65 mM

Assuming temperature = 25.0°C

=( 25 + 273)K

= 298 K

We are to find the concentration of fructose-6-phosphate

Using the relation;

ΔG' = -RT In K_c

where;

R = 8.314 J/K/mol

1670 = - (8.314 × 298 ) In K_c

1670 = -2477.572 × In K_c

1670/ 2477.572 = In K_c

0.67 = In K_c

$K_c = e^{-0.67}$

$K_c =$ 0.511

Now using the equilibrium constant $K_c$

$K_c = \dfrac{[F6P]}{[G6P]}$

$0.511 = \dfrac{[F6P]}{[2.65]}$

F6P = 0.511 × 2.65

F6P = 1.35415

F6P ≅ 1.35 mM

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2 years ago