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stepan [7]
2 years ago
11

How many moles of carbon dioxide can be formed by the decomposition of 5.0 moles of aluminium carbonate

Chemistry
1 answer:
Ksenya-84 [330]2 years ago
3 0

Answer:

30 moles of CO₂

Explanation:

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Give the o.N. Of each of the elements magnesium and oxygen in the reactants and in the products 2Mg + O2=2MgO
vlada-n [284]

Answer:

2Mg^0 + O_2^0\rightarrow2Mg^{2+}O^{2-}

Explanation:

Hello there!

In this case, according to the rules for the oxidation states in chemical reactions, it is possible to realize that lone elements have 0 and since magnesium is in group 2A, it forms the cation Mg⁺² as it loses electrons and oxygen is in group 6A so it forms the anion O⁻²; therefore resulting oxidation numbers are:

2Mg^0 + O_2^0\rightarrow2Mg^{2+}O^{2-}

Best regards!

4 0
3 years ago
Which of these figures probably shows a sugar solution containing a small amount of sugar dissolved in water
Vesnalui [34]

Answer:

The correct answer is Sugar has a greater solubility than sand.

Sugar will easily dissolve in water because it has a lower density than sand, therefore it has a greater solubility.

8 0
4 years ago
Read 2 more answers
Can someone help me I would really appreciate it
slega [8]
The condensed formula would be CH3-CH(CH4)-CH2-CH(CH4)-CH2-CH(CH4)-CH3. The molecular formula would be C10H25.
8 0
4 years ago
How many grams of aluminum oxide (Al2O3) will be formed from 10.0 grams of aluminum (Al)? 4 AL + 3 02 --> 2 AL203​
d1i1m1o1n [39]

Answer:

Mass = 18.9 g

Explanation:

Given data:

Mass of Al₂O₃ formed = ?

Mass of Al = 10.0 g

Solution:

Chemical equation:

4Al + 3O₂      →       2Al₂O₃

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 10.0 g/ 27 g/mol

Number of moles = 0.37 mol

Now we will compare the moles of Al and Al₂O₃.

                      Al          :          Al₂O₃

                       4           :            2

                     0.37        :         2/4×0.37 = 0.185 mol

Mass of Al₂O₃:

Mass = number of moles × molar mass

Mass = 0.185 mol × 101.9 g/mol

Mass = 18.9 g

4 0
3 years ago
Consider a 0.10 M aqueous benzoic acid, CeHeCOOH. The K benzoic acid. 6.5 x 10 for A) Write a balanced equation that shows the r
7nadin3 [17]

Answer:

a) C6H5COOH + H2O ↔ H3O+  +  C6H5COO-

b) [ H3O+ ] = 2.517 E-3 M

c) pH = 2.599

Explanation:

a) balanced equation:

C6H5COOH + H2O ↔ H3O+  +  C6H5COO-

⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5

mass balance:

0.10 m = [ C6H5COO- ] + [ C6H5COOH ].....(1)

charge balance:

[ H3O+ ] = [ C6H5COO- ] + [ OH- ] .......[ OH- ] : comes from water, it's not significant

⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)

b) (2) in (1):

⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]

⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]

⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5

⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0

⇒ [ H3O+ ] = 2.517 E-3 M

c) pH = - log [ H3O+ ]

⇒ pH = - Log ( 2.517 E-3 )

⇒ pH = 2.599

7 0
3 years ago
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