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4vir4ik [10]
3 years ago
14

What element is located in group 5, period 4?​

Chemistry
1 answer:
djverab [1.8K]3 years ago
3 0

Answer:

vanadium

Explanation:

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which field or study (organic chemistry or inorganic chemistry) includes the study of the properties and preparation of chalk?
jeka57 [31]

Answer:the answer is d

Explanation:

3 0
3 years ago
By using the following reactions, calculate the heat of formation of pentane (C5H12) gas from carbon and hydrogen gas. C (s) + O
Llana [10]

Answer: ΔH = 145 kJ/mol


Explanation:


1) Equations given:


C (s) + O₂ (g) → CO₂ (g), ΔH = 395 kJ/mol


H₂ (g) + 1/2O₂ (g) → H₂O (l), ΔH = 285 kJ/mol


C₅H₁₂ (g) + 8O₂ → 5CO₂ (g) + 6H₂O (l), ΔH = 3.54 × 10³ kJ/mol


2) Target equation:


5C(s) +6 H₂(g) → C₅H₁₂(g)


3) Hess law


As per Hess law you can handle the chemical equations (add, subtract, reverse, multiply by a coefficient) to find the target equation, and then the net sum of the heat changes of the individual reactions equals the heat change of the target reaction.


4) Now, you must figure out how to transform the individual reactions to end up with the final equation:


Here is how:


i) Multiply the first one by 5:


5C (s) + 5O₂ (g) → 5CO₂ (g), ΔH = 5×395 kJ/mol


ii) Multiply the second by 6:


6H₂ (g) + 3O₂ (g) → 6H₂O (l), ΔH = 6×285 kJ/mol


iii) Reverse the third:


5CO₂ (g) + 6H₂O (l) → C₅H₁₂ (g) + 8O₂ , ΔH = - 3.54 × 10³ kJ/mol


iv) Add them up:


5C (s) + 5O₂ (g) → 5CO₂ (g), ΔH = 5×395 kJ/mol


6H₂ (g) + 3O₂ (g) → 6H₂O (l), ΔH = 6×285 kJ/mol


5CO₂ (g) + 6H₂O (l) → C₅H₁₂ (g) + 8O₂ , ΔH = - 3.54 × 10³ kJ/mol

----------------------------------------------------------------------------------------------

5C (s) + 6H₂ → C₅H₁₂ (g),


ΔH = 5×395 kJ/mol + 6×285 kJ/mol - 3.54 × 10³ kJ/mol


⇒ ΔH = 145 kJ/mol

6 0
4 years ago
Read 2 more answers
Why doesnt it exist?
belka [17]

Answer:

He2 molecule contains 4 electrons. Each atom gives 2 electrons in 1s orbitals. This way 2 (1s) orbitals combine to give 2 molecular orbitals viz. ... This indicates that there is no bond formation between 2 HE atoms and hence the He2 molecule does not exist.

Explanation:

8 0
4 years ago
The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10
Sati [7]

Answer:

a. 113 min

Explanation:

Considering the equilibrium:-

                   2N₂O₅ ⇔ 4NO₂ + O₂

At t = 0        125 kPa

At t = teq     125 - 2x      4x        x

Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x

125 - 3x = 176 kPa

x = 17 kPa

Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 2.8\times 10^{-3} min⁻¹

Initial concentration [A_0] = 125 kPa

Final concentration [A_t] = 91 kPa

Time = ?

Applying in the above equation, we get that:-

91=125e^{-2.8\times 10^{-3}\times t}

125e^{-2.8\times \:10^{-3}t}=91

-2.8\times \:10^{-3}t=\ln \left(\frac{91}{125}\right)

t=113\ min

3 0
3 years ago
Using the half-life graph of cobalt-57, choose ALL the statements that are accurate.
Morgarella [4.7K]

Answer: B) The half-life of Co-57 is 10 days

D) After twenty days have elapsed, about 25% of the Co-57 remains.

E) Scientists have an 8 gram sample of Co-57. After thirty days, 1 gram remains.

Explanation:

5 0
3 years ago
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