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katrin2010 [14]
3 years ago
14

PLZ HELP! An electron moved from a lower energy level to a higher energy level. What most likely happened during the transition?

Chemistry
1 answer:
mina [271]3 years ago
7 0
Option b) a fixed amount of energy is absorbed
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What amount of water is formed when 20 ml of 0.80 m hcl and 30 ml of 0.40 m naoh are mixed?
Yuri [45]

Answer: The amount of water formed is 12 moles

Explanation: Please see the attachments below

7 0
3 years ago
When a certain powder is mixed into a solution of silver nitrate, a white precipitate forms. When the powder is added to a solut
atroni [7]

Answer:

Ammonium chloride

Explanation:

The powder is:- Ammonium chloride

When mixed with silver nitrate, white prescipitate of silver chloride is formed as:-

NH_4Cl+AgNO_3\rightarrow NH_4NO_3+AgCl

When mixed with sodium hydroxide, ammonia gas is formed which has noxious order.

NH_4Cl +NaOH\rightarrow NH_3 + H_2O + NaCl

Ammonia gas on reaction with nickel (II) hydroxide forms deep blue colored complex as shown below:-

Ni(OH)_2(s) + 6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}(aq) + 2OH^{-}(aq)

5 0
2 years ago
Given the following reaction:
mojhsa [17]

Answer:

Option D. 17.5

Explanation:

Equiibrium is: CO + 2H₂  ⇄  CH₃OH

1 mol of CO is in equibrium with 2 moles of hydrogen in order to make, methanol.

Initially we have 0.42 moles of CO and 0.42 moles of H₂

If 0.29 moles of CO remained, (0.42 - 0.29) = 0.13 moles have reacted.

So in the equilibrium we may have:

0.29 moles of CO, and (0.42 - 0.13 . 2) = 0.16 moles of H₂

Ratio is 1:2, if 0.13 moles of CO haved reacted, (0.13 . 2) moles have reacted of hydrogen

Finally 0.13 moles of methanol, are found after the equilibrium reach the end.

Let's make expression for KC: [Methanol] / [CO] . [Hydrogen]²

0.13 / (0.29 . 0.16²)

Kc = 17.5

4 0
2 years ago
Ethanol melts at -114 degree C. The enthalpy of fusion
Brut [27]

Answer: The heat required is 6.88 kJ.

Explanation:

The conversions involved in this process are :

(1):ethanol(s)(-135^0C)\rightarrow ethanol(s)(-114^0C)\\\\(2):ethanol(s)(-114^0C)\rightarrow ethanol(l)(-114^0C)\\\\(3):ethanol(l)(-114^0C)\rightarrow ethanol(l)(-50^0C)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]

where,

\Delta H = enthalpy change = ?

m = mass of ethanol = 25.0 g

c_{p,s} = specific heat of solid ethanol= 0.97 J/gK

c_{p,l} = specific heat of liquid ethanol = 2.31 J/gK

n = number of moles of ethanol = \frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}}=\frac{25.0g}{46g/mole}=0.543mole

\Delta H_{fusion} = enthalpy change for fusion = 5.02 KJ/mole = 5020 J/mole

T_{final}-T_{initial}=\Delta T = change in temperature

The value of change in temperature always same in Kelvin and degree Celsius.

Now put all the given values in the above expression, we get

\Delta H=[25.0 g\times 0.97J/gK\times (-114-(-135)K]+0.534mole\times 5020J/mole+[25.0g\times 2.31J/gK\times (-50-(-114))K]

\Delta H=6885.93J=6.88kJ     (1 KJ = 1000 J)

Therefore, the heat required is 6.88 kJ

3 0
2 years ago
Like all equilibrium constants, the value of kw depends on temperature. at body temperature (37∘c), kw=2.4⋅10−14. part a what is
LenaWriter [7]
First, we have to know that we can put [H+] instead of [H3O+]

so, according to the reaction equation:

by using ICE table:

             H2O ↔  H+  +  OH-

initial                    0           0

change                +X         +X

Equ                       X             X

when Kw = [H+] [OH-]

and when we have Kw = 2.4 x 10^-14 

and when [H+] = [OH-] = X

∴ 2.4 x 10^-14 = X^2

∴ X = √(2.4 x 10^-14)

      = 1.55 x 10^-7

∴[H+] = 1.55 x 10^-7


6 0
3 years ago
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