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joja [24]
2 years ago
6

What does the chemical formula CaCl, show about the compound it represents? It is made up of one element. It is made up of two e

lements. It is made up of three elements. It is made up of four elements.​

Chemistry
1 answer:
zavuch27 [327]2 years ago
5 0

Answer:

It is made up of two elements.

Explanation:

To answer the question given above,

We shall determine the number of elements present in CaCl₂. This can be obtained as follow:

CaCl₂ contains calcium (Ca) and chlorine gas (Cl₂).

This implies that CaCl₂ contains two different elements.

Now, considering the options given in the question above, CaCl₂ is made up of two elements.

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Answer:

D

Explanation:

The valence electrons are the outer electrons and must be located in the outermost shell. In this case, D.

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3 years ago
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A bomb calorimeter has a heat capacity of 783 J/oC and contains 254 g of water whose specific heat capacity is 4.184 J/goC. How
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Answer : The amount of heat evolved by a reaction is, 4.81 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

q=[q_1+q_2]

q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the water

c_1 = specific heat of calorimeter = 783J/^oC

c_2 = specific heat of water = 4.184J/g^oC

m_2 = mass of water = 254 g

\Delta T = change in temperature = T_2-T_1=(23.73-26.01)=-2.28^oC

Now put all the given values in the above formula, we get:

q=[(783J/^oC\times -2.28^oC)+(254g\times 4.184J/g^oC\times -2.28^oC)]

q=-4208.28J=-4.81kJ

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7 0
3 years ago
Consider the second-order reaction:
kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

Explanation:

2HI(g)\rightarrow H_2(g)+I_2(g)


Rate Law: k[HI]^2


Rate constant of the reaction = k = 6.4\times 10^{-9} L/mol s

Order of the reaction = 2

Initial rate of reaction = R=1.6\times 10^{-7} Mol/L s

Initial concentration of HI =[A_o]

1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2

[A_o]=5 mol/L

Final concentration of HI after t = [A]

t = 4.53\times 10^{10} s

Integrated rate law for second order kinetics is given by:

\frac{1}{[A]}=kt+\frac{1}{[A_o]}

\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}

[A]=0.00345 mol/L

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

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