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WINSTONCH [101]
3 years ago
15

If anyone could please help me i'd greatly appreciate it! I have tried over and over again and can not figure this out.

Chemistry
1 answer:
Papessa [141]3 years ago
7 0

Answer:

%Ionization = 1.63%

Explanation:

Hydrazine in aqueous media theoretically forms a difunctional hydroxyl system. However, for this problem assume only monofunctional ionization occurs. A second hydroxyl ionization would not likely occur as the formal cationic charge formed in the 1st ionization would inhibit a second ionization.

H₂NNH₂ + 2H₂O => HONHNHOH => HONHNH⁺ + OH⁻; Kb = 1.3 x 10⁻⁶

So, assuming all OH⁻ and HONHNH⁺ are delivered in the 1st ionization then a good estimate of the %ionization can be calculated.

               HONHNHOH => HONHNH⁺  +  OH⁻

C(i) =>              0.490M                  0M           0M

ΔC =>                  -x                         +x             +x

C(eq) =>         0.490 - x                    x               x

                      ≅0.490M* => *x is dropped as Conc H₂NNH₂/Kb > 100

Kb = [HONHNH⁺][OH⁻]/[HONHNHOH]

1.3 x 10⁻⁶ = x²/0.490

=> x = [OH⁻] = [HONHNH⁺] = √[(1.3 x 10⁻⁶)(0.490)] = 8 x 10⁻⁴

=> %Ionization = (x/0.490)100% = (8 x 10⁻⁴/0.490)100% = 1.63%

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3 years ago
At a festival, spherical balloons with a radius of 170.cm are to be inflated with hot air and released. The air at the festival
MissTica

Answer:

The correct answer is 1.5 balloons.

Explanation:

Based on the given question, the radius of the spherical balloons is 170 cm. Therefore, the volume of each balloon will be,  

= 4/3 × π × (170)³ cm³

= 20.5176 × 10⁶ cm³  

= 20.5176 m³

The density of air at 100 degree C s 0.946 Kg m⁻³

The mass of air in each balloon can be calculated by using the formula,  

Mass = density × volume

Mass = 0.946 Kg m⁻³ × 20.5176 m³

Mass = 19.410 Kg

The heat energy, that is, required to bring the air from 25 degree C to 100 degree C will be,  

= 19.410 × 10³ g × (100 -25) degree C × 1.009 J/g degree C

= 14.68 × 10⁵ J

The concentration of butane given is 1.00 Kg or 1000 g

The molecular weight of butane is 58.12 g per mole

The moles or n can be calculated by using the formula,  

n = mass / mol.wt

n = 1000 g/58.12 g/mol = 17.20 mol

The formation of enthalpy of butane at 25 degree C is 125.7 × 10³ J/mol. The evolution of heat energy that take place at the time of burning 17.20 moles of butane is,  

= 125.7 × 10³ J/mol × 17.20 mol

= 2.16 × 10⁶ J

The number of balloons that can be inflated with hot air is,  

= 2.16 × 10⁶ J / 14.68 × 10⁵ J per each balloon

= 1.5 balloons

Hence, maximum of 1.5 balloons can be inflated.  

7 0
3 years ago
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