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Kazeer [188]
3 years ago
5

What is the molecular formula of the hydrocarbon whose molar mass is 536 g/mol and contains 89.55 w/w% carbon?

Chemistry
1 answer:
leonid [27]3 years ago
7 0

Given the percentage composition of HC as C → 81.82 % and H → 18.18 %

So the ratio of number if atoms of C and H in its molecule can will be:

C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8

So the Empirical Formula of hydrocarbon is:

C 3 H 8

As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol

Now let Molecular formula of the HC be ( C 3 H 8 ) n

Using molar mass of C and H the molar mass of the HC from its molecular formula is:

( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1

Hence the molecular formula of HC is C 3 H 8

Does that help?

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Show the difference in the reactivity of Na,Ca,Mg,Al,Fe,Cu with H2O
Vesnalui [34]

Answer:

Explanation:

Na react with H2O to form NAOH

2 Na+2H2O....................2NAOH + H2

Ca react with water and form calcium hydroxide

Ca + 2H2O........................Ca(OH)2

Mg react with water and form Magnesium hydroxide

Mg +2H2O .........................Mg(OH)2 however this coating of mg(oh)2 prevent it from further reaction

 Fe react with water and form ferric hydride

3Fe +H2O.......................2 FeH +FeO

copper do not react with water

7 0
3 years ago
The inputs for cellular respiration are
katrin [286]
Glycolysis--The breakdown of a glucose molecule into two three-carbon pieces called pyruvate. You will notice that very little ATP is produced in this step and no oxygen is required. ... This step is also where other molecules besides glucose may be fed into the cell respiration<span> process, especially lipids.</span>
5 0
3 years ago
Read 2 more answers
Which type of reactions form salts?
s344n2d4d5 [400]

Neutralization reactions are the reactions type which form salts.

Explanation:

Salts are formed by ionic bonds when the oxidation states of anions and cations are equal and have opposite signs. So one should be highly electronegative in nature and another should be highly electropositive in nature. So the electropositive element will be ready to give electrons and the electronegative element will be ready to accept all the electrons given by the electropositive element. As a whole the compound will be neutrally charged by adding of equal number of positively charged and negatively charged ions.

The reduction or addition of electrons will be occurring in cations and the oxidation or removal of electrons will be occurring in anions.

So the salt formation is based on neutralization reactions.

8 0
3 years ago
If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W
olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3

Therefore, the leftover of sodium nitrate is:

m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3

Finally, the percent yield is computed via:

Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%

Best regards!

6 0
2 years ago
*CHEMISTRY*
Troyanec [42]

Answer:

1.54 liters.

Explanation:

If the liters increases by .27 for every 100ºC, then just multiply .27 by 2.

You'd then get 1.54, which is your answer.

Hope this helps!

8 0
2 years ago
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