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Finger [1]
3 years ago
10

Why does a 4 carbon linear alkane, butane c4h10, have two more hydrogens ?

Chemistry
1 answer:
Scilla [17]3 years ago
3 0
Alkanes are hydrocarbons that only contain single bonds in them. A carbon can bond with up to 4 atoms, even with another carbon atom. So, in a C-C bond, 3 more H atoms can bond to each of the C atom. Generally, the chemical formula for alkanes is CₓH₂ₓ₊₂. So for butane, there are 4 C atoms. The corresponding H atoms are 2(4) + 2 = 10. That's why it's chemical formula is C₄H₁₀.
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xeze [42]

Answer:

5 atoms are in each molecule of calcium carbonate

6 0
3 years ago
How many energy levels does the element helium have?
Alex Ar [27]

Answer:

Element Helium have<u> 2 energy levels. </u>

Explanation:

<u>HELIUM -</u> With the symbol He and the atomic number 2, helium is a chemical element. It is a colorless , odorless, tasteless, non-toxic, inert, monatomic gas that is the first in the periodic table in the noble gas group. Of all the elements, its boiling point is the lowest.

At about 24 % of the total elemental mass, which is more than 12 times the mass of all the heavier elements combined, it is present. In both the Sun and Jupiter, its abundance is identical to this. This is due to helium-4's very high nuclear binding energy (per nucleon) with respect to the next three elements following helium.

<u>Hence , the answer is 2 energy levels in Helium atom.</u>

8 0
3 years ago
A scientist claims to have a cooling apparatus kept at -100C by liquid nitrogen. Is this possible? Why or why not?
Vilka [71]

Answer:

It is not possible

Explanation:

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5 0
4 years ago
What are atoms help
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8 0
3 years ago
Al2(SO4)3(s) + H2O(l) ---- Al2O3(s) + H2SO4 (aq) calculate enthalpy formation for this reaction. Balance the reaction. Calculate
katen-ka-za [31]

The given chemical equation is:

Al_{2}(SO_{4})_{3}(aq)+H_{2}O(l)-->Al_{2}O_{3}(aq)+H_{2}SO_{4}(aq)

On balancing the equation we get,

Al_{2}(SO_{4})_{3}(aq)+3H_{2}O(l)-->Al_{2}O_{3}(aq)+3H_{2}SO_{4}(aq)

Calculating enthalpy of formation of this reaction from the standard heats of formation of the products and reactants:

ΔH_{reaction}^{0}=[H_{f}^{0}(Al_{2}O_{3}(s)) + (3*H_{f}^{0}(H_{2}SO_{4}(aq))] -   [H_{f}^{0}(Al_{2}SO_{4}(aq)) + (3*H_{f}^{0}(H_{2}O(l))]

=[(-1669.8kJ/mol)+ {3* (-909.27 kJ/mol)}]-[(-3442kJ/mol)+{3*(-285.8 kJ/mol)}]

=[(-4397.61kJ/mol)]-[(-4299.4kJ/mol)]

=-98.21kJ/mol

Total enthalpy change when 15 mol of Al_{2}(SO_{4})_{3}reacts will be=

15 mol Al_{2}(SO_{4})_{3}*\frac{-98.21kJ}{1 molAl_{2}(SO_{4})_{3}} =-1473.15kJ/mol

4 0
3 years ago
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