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Olegator [25]
3 years ago
11

AlCl3 + Na NaCl + Al Did Cl change oxidation number?

Chemistry
2 answers:
gizmo_the_mogwai [7]3 years ago
6 0
Yes. If this is the balanced equation:

AlCl3 + 3Na —— 3NaCl + Al

then Al was reduced from a 3+ oxidation (to neutralize the 3- from the chlorine) to a 0 oxidation (elemental ground state).
damaskus [11]3 years ago
5 0

Answer : There is no change takes place in the oxidation number of Cl.

Explanation :

The given chemical reaction is,

AlCl_3+Na\rightarrow NaCl+Al

This reaction is an unbalanced reaction because the chlorine atoms are not balanced.

In order to balance the chemical reaction, the coefficient 3 is put before the Na and NaCl.

The balanced chemical reaction will be,

AlCl_3+3Na\rightarrow 3NaCl+Al

Now we have to calculate the oxidation number of all the elements.

In AlCl_3, the oxidation number of Al and Cl are, (+3) and (-1) respectively.

The oxidation number Na and Al are, zero (0)

In NaCl, the oxidation number of Na and Cl are, (+1) and (-1) respectively.

From this we conclude that the oxidation number of Cl changes from (-1) to (-1) that means remains same.

Therefore, there is no change takes place in the oxidation number of Cl.

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Pavel [41]
<span>Since,
1000 grams of water = 1000 mL of water</span><span>
So, 
At any of the given temperature:
</span>1000 mL = 10 x 100 mL
<span>
moles of NH4Cl = 53.5/53.49
                          = 1.0 m
                          = 1.0 mol/Kg
Delta T = 2 x 1.86 x 1.0
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3 0
3 years ago
A 1.0-L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys the buffer? (a) adding 0.050 mol of HCl (b) ad
Volgvan

Answer:

(a) adding 0.050 mol of HCl

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.

In the buffer:

1.0L × (0.10 mol / L) = 0.10 moles of HF -<em>Weak acid-</em>

1.0L × (0.050 mol / L) = 0.050 moles of NaF -<em>Conjugate base-</em>

-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-

Thus:

<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.

(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.

(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>

3 0
3 years ago
Reactants are substances that?
RSB [31]
A substance that undergoes change during a reaction, usually from coming in contact with another substance
4 0
3 years ago
Which of the following statements is true?
frez [133]
Answer: D

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3 0
2 years ago
The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:                                                        

k = Ae^{-\frac{E_{a}}{RT}}

For the reaction without catalyst we have:

k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}   (1)

And for the reaction with the catalyst:

k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}   (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:                      

\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}

\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}    

\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}    

Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

And assuming that the initial concentrations ([R]) are the same, we have:

\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

4 0
3 years ago
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