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o-na [289]
3 years ago
12

What is the similarity between radioactive iodine and stable iodine

Chemistry
1 answer:
Elena-2011 [213]3 years ago
7 0

Answer:

both iodine

Explanation:

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A gas mixture is composed of three different gases. The first gas in the mixture has a pressure of 0.35 atm. The second gas in t
riadik2000 [5.3K]

Answer:

1.20atm

Explanation:

Given parameters:

Partial pressure of gas 1 = 0.35atm

Partial pressure of gas 2  = 0.20atm

Partial pressure of gas 3  = 0.65atm

Unknown:

Total pressure of the gas mixture = ?

Solution:

To solve this problem, we need to recall and understand the Dalton's law of partial pressure.

Dalton's law of partial pressure states that "the total pressure of a mixture of gases is equal to the sum of the partial pressure of the constituent gases".

    Total pressure  =Pressure of gas(1 + 2 + 3)

The partial pressure is the pressure a gas would exert if it alone occupied the volume of the gas mixture.

     

Now we substitute;

      Total pressure  = (0.35 + 0.20 + 0.65)atm  = 1.20atm

8 0
3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
How is a compound different from a mixture? *
andreyandreev [35.5K]

Answer:

a.  mixtures are created through physical changes and compounds are created through chemical reactions

Explanation:

5 0
3 years ago
Briefly explain how magnetic behavior is related to the arrangement of valence electrons in the outer orbitals
olga nikolaevna [1]

Answer:

Elements with unpaired valence electrons in the outer orbital are paramagnetic. Elements that are predicted to exhibit paramagnetic behavior may exhibit ferromagnetic behavior. Elements that contain only paired valence electrons are diamagnetic.

3 0
3 years ago
A chemical company makes two brands of antifreeze. the first brand is 70% pure antifreeze, and the second brand is 95% pure anti
madreJ [45]

Answer is: 56 gallons of 70% antifreeze and 84 gallons of 95% antifreeze.


ω₁ = 70% ÷ 100% = 0.7; 70% pure antifreeze.

ω₂ = 95% ÷ 100% = 0.95.

ω₃<span> = 85% ÷ 100% = 0.85.
V</span>₁ = ?; volume of 70% antifreeze.

V₂ = ?; volume of 95% antifreeze.<span>
V</span>₃ = V₁ + V₂<span>.
V</span>₃ = 140 gal.

V₁ = 140 gal - V₂<span>.
ω</span>₁ · V₁ + ω₂ ·V₂ = ω₃ · V₃.

0.70 · (140 gal - V₂) + 0.95 · V₂ = 0.85 · 140 gal.

98 gal - 0.7V₂ + 0.95V₂ = 119 gal.

0.25V₂ = 21 gal.

V₂ = 21 gal ÷ 0.25.

V₂ = 84 gal.

V₁ = 140 gal - 84 gal.

V₁ = 56 gal.

4 0
3 years ago
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