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nekit [7.7K]
3 years ago
12

Example 13:0.29 grams of a hydrocarbon with vapour density 29 when burnt completely in oxygen produce 448 ml of carbon dioxide a

t S.T.P. From the given information, calculate the (i) mass of carbon dioxide formed.​
Chemistry
1 answer:
Juliette [100K]3 years ago
6 0

Answer:

Given, 0.29 g of hydrocarbon produces 448ml of CO2 at STP. then, C2H5 is the emperical formula of hydrocarbon . n = 2 , hence, molecular formula will be C4H10

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What is the ph of a solution that results from mixing 25 ml of 0.15 M hcl to 25 ml of 0.52 m nh3?
BartSMP [9]

Answer:

7.00

Explanation:

When the solutions are mixed, the HCl dissociates to form the ions H+ and Cl-. The ion H+ will react with the NH3 to form NH4+. The stoichiometry for this is 1 mol of HCl to 1 mol of H+ to 1 mol of Cl-, and 1 mol of H+ to 1 mol of NH3 to 1 mol of NH4+.

First, let's find the number of moles of each one of them, multiplying the concentration by the volume:

nH+ = 0.15 M * 25 mL = 3.75 mmol

nNH3 = 0.52 M * 25 mL = 13 mmol

So, all the H+ is consumed, and the neutralization is completed, thus pH will be the pH of the solvent (water), pH = 7.00.

5 0
3 years ago
How has the fishing industry along the Chesapeake Bay changed in the last 25 years?
zhuklara [117]

Answer:

In the last 25 years, the fish hunt in the Chesapeake Bay occurred very fastly due to which the fish industry enhanced. However, this rapid fish industrialization caused many of the fish species to become endangered. Hence, the fish industries started to use two basic management techniques which were conservation and allocation.

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3 years ago
A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent
Degger [83]

Answer :  The % of (+) limonene isomer = 79%


                The % of (-) limonene isomer = 0%


                The % of enantiomeric excess = 58%


Explanation :   Enantiomeric excess (ee) is the measurement of purity used for chiral substances.


Given,


% of pure limonene enantiomer = The % of (+) limonene isomer = 79%


Therefore, The % of (-) limonene isomer = 0%


Formula used :  

\%(+)\text{ isomer}=\frac{ee}{2}+50\%


Where,         ee → enantiomeric excess


Now, put all the values in above formula, we get the value of enantiomeric excess (ee).


     {ee}=\frac{\%(+)-50\%}{2}


            =\frac{79\%-50\%}{2}


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7 0
3 years ago
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the standard change in Gibbs free energy is Δ????°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate Δ????ΔG for this reaction at 298 K29
Vera_Pavlovna [14]

Answer:

ΔG = 16.218 KJ/mol

Explanation:

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∴  R = 8.314 J/K.mol

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