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Kazeer [188]
3 years ago
10

D. liquid

Chemistry
1 answer:
erastovalidia [21]3 years ago
4 0

Answer: H2O

Explanation: The given balanced chemical reaction is,

This reaction is a reversible reaction.

The rate of forward reaction will be,

The rate of backward reaction will be,

And at equilibrium the rate of reaction is equal to the rate of backward reaction divided by the rate of forward reaction.

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Which of the following results in the release of nuclear energy?
Pepsi [2]

c.Both the breaking of nuclear bonds and the forming of nuclear bonds.

Explanation:

Nuclear energy is released by the breaking and forming of nuclear bonds. The breaking of nuclear bonds by unstable atoms is known as nuclear fission. The forming of nuclear bonds by combination of light atoms is known as nuclear fusion.

  • Nuclear fission is a radioactive decay process in which a heavy nucleus spontaneously disintegrates into lighter ones with the release of energy.
  • In nuclear fusion, atomic nuclei combines into larger ones with the release of large amount of energy.

Learn more:

Nuclear decay brainly.com/question/4207569

#learnwithBrainly

3 0
3 years ago
Calculate the moles in 45.06g of Be
g100num [7]
5.00111 moles in 45.05g of br
5 0
3 years ago
Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M
Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

Half reactions for the given cell follows:

Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

5 0
4 years ago
The initial concentration of reactant in a first-order reaction is 0.27 m. the rate constant for the reaction is 0.75 s-1. what
hjlf
The concentration in  mol/l  of reactant  after  1.5 s  is calculated as  follows

from  first order  integrated  equation

 In (A)t  = - Kt + In (A)o  where
  At = final  concentration =?
A)o =initial  concentration =0.27 M
K =  constant=0.75
T=temperature =  1.5 s

in( A)t = -0.75(1.5) + in(0.27)

In (A)t =-1.125  +-1.31 =-2.435
In(A)t =-2.435

In (A)t  =  e
find the e  value 

 ( A)t is therefore =0.0876  mol/l

5 0
4 years ago
Read 2 more answers
How much momentum does a 200 kg rhino have that is running at 35 m/s E
nirvana33 [79]

Answer:

7000 kg*m/s E

Explanation:

Momentum formula: p=mv

m=200kg

v=35 m/s East

p=(200kg)(35m/s E)

m=7000 kg*m/s E

If you want to simplify it further, m=7*10^3 kg*m/s E

5 0
3 years ago
Read 2 more answers
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