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2 years ago

7

Which of the following chemical reactions is the only reaction shown that would be predicted to occur based on the activity seri

es?
A. Ca + BaBr2- Ba + CaBr2
B. Mn + NiBr2 - Ni + MnBr2
C. Au + PtBr - Pt + AuBr
D. Zn + MgBr2 Mg + ZnBr2

1 answer:

Novosadov [1.4K]2 years ago

7 0

B. Mn + NiBr₂ → Ni + MnBr₂

Explanation:

The reaction that can be predicted of all is Mn + NiBr₂ → Ni + MnBr₂.

The activity series is used to predict the products of single displacement reactions.

The series ranks metals in order of their reactivity. Those higher up in the series are highly reactive metals. Those at the bottom are slightly to non-reactive metals.

For a single displacement reaction to occur, a metal higher up in the activity series displaces one that is lower in the series.

Reaction A will not occur, Ba is higher in the series

Reaction C will not occur, Pt and Au are unreactive

Reaction D will not occur as Zn is lower in the series

Mn is higher in the reactivity series and it will displace Ni from the solution.

Learn more:

Synthesis reaction brainly.com/question/4216541

#learnwithBrainly

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Which distinguishes an atom of one element from an atoms of a different element?

dalvyx [7]

Answer:

The number of protons you welcome

Explanation:

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3 years ago

If 45.0 mL of a 0.0500 M HNO3, 10.0 mL of a 0.0500 M KSCN, and 30.0 mL of a 0.0500 M Fe(NO3)3 are combined, what is the initial

jeka94

Answer:

the initial concentration of SCN- in the mixture is 0.00588 M

Explanation:

The computation of the initial concentration of the SCN^- in the mixture is as follows:

As we know that

$KSCN \rightarrow K^ + SCN^-$

As it is mentioned in the question that KSCN is present 10 mL of 0.05 M

So, the total milimoles of SCN^- is

= 10 × 0.05

= 0.5 m moles

The total volume in mixture is

= 45 + 10 + 30

= 85 mL

Now the initial concentration of the SCN^- is

= 0.5 ÷ 85

= 0.00588 M

hence, the initial concentration of SCN- in the mixture is 0.00588 M

5 0

3 years ago

How many moles of oxygen are present in 33.6l of the gas at 1atm and 0c

andreev551 [17]

The answer is: 1.5 moles of oxygen are present.

V(O₂) = 33.6 L; volume of oxygen.

p(O₂) = 1.0 atm; pressure of oxygen.

T = 0°C; temperature.

Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).

At STP one mole of gas occupies 22.4 liters of volume.

n(O₂) = V(O₂) ÷ Vm.

n(O₂) = 33.6 L ÷ 22.4 L/mol.

n(O₂) = 1.50 mol; amount of oxygen.

5 0

3 years ago

An increase in temperature results in A) a decrease in the required activation energy while the reaction rate remains constant.

N76 [4]

Answer:

C) an increase in rate of reaction because reactant molecules collide with greater energy

Explanation:

Temperature is one of the factors that affect the rate of a reaction. The rate of a reaction increases with an increase in temperature and vice versa. When the temperature of a reaction increases, the kinetic energy of the reactant molecules increases causing them to react at a faster rate.

The reactant molecules respond to an increase in temperature by colliding at a faster rate due to an increased kinetic energy between the reactant molecules.

7 0

3 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago