The new temperature (in °C) of the gas, given the data is –148.20 °C
<h3>Data obtained from the question </h3>
- Initial temperature (T₁) = 149.05 °C = 149.05 + 273 = 422.05 K
- Initial pressure (P₁) = 349.84 KPa
- Volume = constant
- New pressure (P₂) = 103.45 KPa
- New temperature (T₂) =?
<h3>How to determine the new temperature </h3>
The new temperature of the gas can be obtained by using the combined gas equation as illustrated below:
P₁V₁ / T₁ = P₂V₂ / T₂
Since the volume is constant, we have:
P₁ / T₁ = P₂ / T₂
349.84 / 422.05 = 103.45 / T₂
Cross multiply
349.84 × T₂ = 103.45 × 422.05
Divide both side by 349.84
T₂ = (103.45 × 422.05) / 349.84
T₂ = 124.80 K
Subtract 273 from 124.80 K to express in degree celsius
T₂ = 124.80 – 273
T₂ = –148.20 °C
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When a solid forms with two solutions are mixed it is a precipitate
Answer:
m = 4450 g
Explanation:
Given data:
Amount of heat added = 4.45 Kcal ( 4.45 kcal ×1000 cal/ 1kcal = 4450 cal)
Initial temperature = 23.0°C
Final temperature = 57.8°C
Specific heat capacity of water = 1 cal/g.°C
Mass of water in gram = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 57.8°C - 23.0°C
ΔT = 34.8°C
4450 cal = m × 1 cal/g.°C × 34.8°C
m = 4450 cal / 1 cal/g
m = 4450 g
The <span>covalent bonds are predicted for each atom are :
</span>(a)F = 1
(b) Si = 4
(c) Br = 1
(d) O = 2
(e) P = 3
(f) S = 2
The answer to your question is,
B. Government.
-Mabel <3
