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MAXImum [283]
3 years ago
6

NEED HELP! Please refer to the picture for the question!

Chemistry
1 answer:
Mashutka [201]3 years ago
4 0

Answer: i think the best bet i can give you is Option C (2:3)

Explanation: i apologize i haven't done chem in 2 years

but however to put it in retrospect the finished equation is 2(AL)^+3 3(O)^-2

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What is chemical energy?
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Answer:
I think it’s D
Explanation:
Chemical strength, Chemical compounds have energy contained in their bonds. Chemical energy may be emitted in the form of heat during a chemical reaction, which is known as an exothermic reaction. The body transforms the chemical energy in food into mechanical energy and heat.
8 0
3 years ago
What is the concentration of a saturated solution of SrSO4? SrSO4 has a Ksp of 3.2 x 10–7
Andrej [43]
SrSo4 = Sr(2+) + SO4(2-)

Let’s say that the initial concentration of SrSo4 was 1. ( or we have 1 mole of this reagent).

When The reaction occurs part of SrSo4is dissociated. And we get X mole Sr(2+) and So4(2-).
Ksp=[Sr(2+)]*[SO4(2-)]
X^2=3.2*10^-7
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5 0
3 years ago
Which is one of the big ideas that guides Earth science?
Nina [5.8K]

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LIFE

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5 0
3 years ago
Which is always going to be the highest ionization energy?
Law Incorporation [45]

Answer: I think it's B.

Explanation:

4 0
2 years ago
A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a
deff fn [24]

Answer:

A) E_{a} = 350KJ/mol, E_{a} = 50KJ/mol, E_{a} = 50KJ/mol

     A = 1.5×10^{-7}s^{-1}, A = 1.9×10^{-7} s^{-1}, A=1.5×10^{-7} s^{-1}

B) 4.469

Explanation:

From Arrhenius equation

      K=Ae^{\frac{E_{a} }{RT} }

where; K = Rate of constant

            A = Pre exponetial factor

            E_{a} = Activation Energy

             R = Universal constant

             T = Temperature in Kelvin

Given parameters:

E_{a} =165KJ/mol

T_{1}=505K

T_{2}=525K

R=8.314JK^{-1}mol^{-1}

taking logarithm on both sides of the equation we have;

InK=InA-\frac{E_{a} }{RT}

since we have the rate of two different temperature the equation can be derived as:

In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )

In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1}  }.(\frac{1}{505} -\frac{1}{525} )

In(\frac{K_{2} }{K_{1} } )= 19846.04×7.544×10^{-5} = 1.497

\frac{K_{2} }{K_{1} } =e^{1.497} = 4.469

 

6 0
3 years ago
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