Answer:
221 °C
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 4.1 L
Initial temperature (T₁) = 25 °C
= 25 °C + 273
= 298 K
Final volume (V₂) = 6.8 L
Final temperature (T₂) =?
The final temperature of the gas can be obtained as follow:
V₁ / T₁ = V₂ / T₂
4.1 / 298 = 6.8 / T₂
Cross multiply
4.1 × T₂ = 298 × 6.8
4.1 × T₂ = 2026.4
Divide both side by 4.1
T₂ = 2026.4 / 4.1
T₂ ≈ 494 K
Finally, we shall convert 494 K to celcius temperature. This can be obtained as follow:
°C = K – 273
K = 494
°C = 494 – 273
°C = 221 °C
Thus the final temperature of the gas is 221 °C
She would observe a yellowish solid precipitate which is the lead iodide and a white solid precipitate which is the potassium nitrate.
This is because the lead nitrate solution which contains particles of lead will mix with the potassium iodide solution containing particles of iodide. Upon mixing,the lead particles from the Lead nitrate solution combines with the iodide particles from Potassium iodide and form two compounds, a yellowish solid precipitate called lead iodide and a white solid precipitate called Potassium nitrate.
The formation of entirely two new compounds is known as the double displacement reaction and can be written in a chemical equation as
2KI(aq.)+Pb(NO₃)₂(aq.)------>2KNO₃(aq.)+PbI ₂(s)
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Around two seconds for an inhalation and three seconds exhalation
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