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STatiana [176]
3 years ago
12

Determine the number of valence electrons for aluminum through the use of this table.

Chemistry
2 answers:
Zarrin [17]3 years ago
7 0

Answer:

Valence of Aluminium: +3

Explanation:

The only known number of valence electrons for the Al combined with other elements is +3. For example in substances such as Al2O3 (alumina).

julia-pushkina [17]3 years ago
3 0
There appears to be no table shown, but I can still answer. Aluminum has 3 valence electrons 
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assume in a different experiment, you prepare a mixture containing 10.0 M FeSCN2+, 1.0 M H+, 0.1 MFe3+ and 0.1 M HSCN. Is the in
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Answer:

The mixture is not in equilibrium, the reaction will shift to the left.

Explanation:

<em>Based on the equilibrium:</em>

<em>Fe³⁺+ HSCN ⇄ FeSCN²⁺ + H⁺</em>

<em>kc = 30 = [FeSCN²⁺] [H⁺] / [Fe³⁺] [HSCN]</em>

Where [] are concentrations at equilibrium. The reaction is in equilibrium when  the ratio of concentrations = kc

Q is the same expression than kc but with [] that are not in equilibrium

Replacing:

Q = [10.0M] [1.0M] / [0.1M] [0.1M]

Q = 1000

As Q > kc, the reaction will shift to the left in order to produce Fe³⁺ and HSCN untill Q = Kc

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7 0
2 years ago
Gas A effuses 0.68 times as fast as Gas B. If the molar mass of Gas B is 17 g, what is the mass of Gas A?
BartSMP [9]
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4 0
3 years ago
Calculate the number of grams of sulfuric acid in 1 gallon of battery acid if the solution has a density of 1.25 g/mL and is 37.
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Answer:

break it down and just put the numbers

Explanation:

4 0
3 years ago
Discuss the different observations that you would record during an investigation into the energy transformations of a lamp that
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Answer:

See the answer below, please.

Explanation:

Take as an example a light bulb inside a lamp to illuminate a room. When you plug it in a plug and turn it on, light is generated. More precisely, heat (Joule effect) is produced inside the lamp by its internal filament (conductive material) when it passes through the electrical energy, generated by the friction of the atoms that are inside it when it encounters a resistance.

3 0
3 years ago
rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is
Sunny_sXe [5.5K]

Complete Question

The  rate of a certain reaction is given by the following rate law:

            rate =  k [H_2][I_2]

rate Use this information to answer the questions below.

What is the reaction order in H_2?

What is the reaction order in I_2?

What is overall reaction order?

At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer:

The reaction order in H_2 is  n =  1

The reaction order in I_2 is  m = 1

The  overall reaction order z =  2

When the hydrogen is double the the initial rate is   rate_n  =  4.0*10^{-4} M/s

The rate constant is   k = 35.23 \  M^{-1} s^{-1}

Explanation:

From the question we are told that

   The rate law is  rate =  k [H_2][I_2]

   The rate of reaction is rate =  2.0 *10^{4} M /s

Let the reaction order for H_2 be  n and for I_2  be  m

From the given rate law the concentration of H_2 is raised to the power of 1 and this is same with I_2 so their reaction order is  n=m=1

   The overall reaction order is  

               z  = n +m

               z  =1 +1

               z  =2

At  rate =  2.0 *10^{4} M /s

        2.0*10^{4}  = k  [H_2] [I_2] ---(1)

= >    k  = \frac{2.0*10^{4}}{[H_2] [I_2]  }

given that the concentration of hydrogen is doubled we have that

            rate  = k [2H_2] [I_2] ----(2)

=>      k = \frac{rate_n  }{ [2H_2] [I_2]}

 So equating the two k

           \frac{2.0*10^{4}}{[H_2] [I_2]  } = \frac{rate_n  }{ [2H_2] [I_2]}

    =>    rate_n  =  4.0*10^{-4} M/s

So when

      rate_x =  52.0 M/s

        [H_2] = 1.8 M

         [I_2] =  0.82 \ M

We have

      52 .0 =  k(1.8)* (0.82)

     k = \frac{52 .0}{(1.8)* (0.82)}

      k = 35.23 M^{-2} s^{-1}

     

     

3 0
3 years ago
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