Calcium carbide ( CaC2) reacts with water to produce acetylene (C2H2) as shown in the unbalanced reaction below: CaC2(s)+H2O(g)-
1 answer:
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Explanation:
The given data is as follows.
Volume of water = 0.25
Density of water = 1000
Therefore, mass of water = Density × Volume
=
= 250 kg
Initial Temperature of water () =
Final temperature of water =
Heat of vaporization of water () at
is 2133 kJ/kg
Specific heat capacity of water = 4.184 kJ/kg/K
As 25% of water got evaporated at its boiling point () 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)
= 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 = = 37 kJ/s or kW
The power rating of electric heating element is 37 kW.
Hence, heat required to raise the temperature from to
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
=
= 3392 sec
Time required to raise the temperature from to
of 0.25
water is calculated as follows.
= 56 min
Thus, we can conclude that the time required to raise the temperature is 56 min.
Answer:
0.1035 M
Explanation:
Considering:
Sodium chloride will furnish Sodium ions as:
Given :
For Sodium chloride :
Molarity = 0.288 M
Volume = 3.58 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 3.58×10⁻³ L
Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:
Moles of sodium ions by sodium chloride = 0.00103104 moles
Sodium sulfate will furnish Sodium ions as:
Given :
For Sodium sulfate :
Molarity = 0.001 M
Volume = 6.51 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 6.51 ×10⁻³ L
Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:
Moles of sodium ions by Sodium sulfate = 0.00001302 moles
Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles
Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L
Concentration of sodium ions is:
<u> The final concentration of sodium anion = 0.1035 M</u>