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storchak [24]
4 years ago
14

Which of the following statement is FALSE? a. Lowering the free energy of the transition state can increase a reaction rate. b.

An increase in temperature can result in an increased reaction rate. c. At a given temperature and time all molecules in a solution or a sample will have the same energy. d. The free energy barrier in a chemical reaction must be overcome in order for products to form.
Chemistry
1 answer:
ryzh [129]4 years ago
8 0

Answer: Option (c) is the correct answer.

Explanation:

Activation energy or free energy of a transition state is defined as the  minimum amount of energy required to by reactant molecules to undergo a chemical reaction.

So, when activation energy is decreased then molecules with lesser amount of energy can also participate in the reaction. This leads to an increase in rate of reaction.  

Also, increase in temperature will help in increasing the rate of reaction.

Whereas at a given temperature, every molecule will have different energy because every molecule travels at different speed.

Hence, we can conclude that out of the given options false statement is that at a given temperature and time all molecules in a solution or a sample will have the same energy.

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A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.
BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\  &=185.59\text{ g/mol} \end{aligned}

Dimensional Analysis:

\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0
3 years ago
You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,
bonufazy [111]

Answer:

\frac{[F^{-}]}{[HF]} is larger

Explanation:

pK_{a}=-logK_{a} , where K_{a} is the acid dissociation constant.

For a monoprotic acid e.g. HA, K_{a}=\frac{[H^{+}][A^{-}]}{[HA]} and \frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}

So, clearly, higher the K_{a} value , lower will the the pK_{a}

In this mixture, at equilibrium, [H^{+}] will be constant.

K_{a} of HF is grater than K_{a} of HCN

Hence, (\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})

So, \frac{[F^{-}]}{[HF]} is larger

5 0
3 years ago
2. Metals react with water and release hydrogen gas. Explain why non-metals do not release hydrogen gas when reacted with water.
Phantasy [73]

Answer:

Its because non-metals are unable to break the bond between the H and O ion and cannot reduce hydrogen by donating electrons

8 0
2 years ago
How many moles of chloride ions are there in 2.5 L of 5 M magnesium chloride?
castortr0y [4]

Answer:

n_{Cl^-}=25molCl^-

Explanation:

Hello,

In this case, since the given 5-M concentration of magnesium chloride is expressed as:

5\frac{molMgCl_2}{L}

We can notice that one mole of salt contains two moles of chloride ions as the subscript of chlorine is two, in such a way, with the volume of solution we obtain the moles of chloride ions as shown below:

n_{Cl^-}=5\frac{molMgCl_2}{L}*\frac{2molCl^-}{1molMgCl_2} *2.5L\\\\n_{Cl^-}=25molCl^-

Best regards.

4 0
3 years ago
What mass of mgo is produced from 2.00 moles of mg ? Mg +02 mgo
Vanyuwa [196]
Chemical Reactions and Moles of Reactants and Products
That is, it requires 2 moles of magnesium and 1 mole of oxygen to produce 2 moles of magnesium oxide. If only 1 mole of magnesium was present, it would require 1 ÷ 2 = ½ mole of oxygen gas to produce 2 ÷ 2 = 1 mole magnesium oxide.
3 0
3 years ago
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