Answer:
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Answer:
The final temperature of water is 54.5 °C.
Explanation:
Given data:
Energy transferred = 65 Kj
Mass of water = 450 g
Initial temperature = T1 = 20 °C
Final temperature= T2 = ?
Solution:
First of all we will convert the heat in Kj to joule.
1 Kj = 1000 j
65× 1000 = 65000 j
specific heat of water is 4.186 J /g. °C
Formula:
q = m × c × ΔT
ΔT = T2 - T1
Now we will put the values in Formula.
65000 j = 450 g × 4.186 J /g. °C × (T2 - 20°C )
65000 j = 1883.7 j /°C × (T2 - 20°C )
65000 j/ 1883.7 j /°C = T2 - 20°C
34.51 °C = T2 - 20°C
34.51 °C + 20 °C = T2
T2 = 54.5 °C
Answer:
The answer to your question is P2 = 84.16 kPa
Explanation:
Data
Volume 1 = V1 = 4.52 L Volume 2 = V2 = 4.83 l
Pressure 1 = P1 = 102 kPa Pressure 2 = P2 = ?
Temperature 1 = T1 = 23°C Temperature 2 = T2 = -12°C
Process
1.- Convert the temperature to °K
Temperature 1 = 23 + 273 = 296°K
Temperature 2 = -12 + 273 = 261°K
2.- Use the Combined Gas law to solve this problem
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = P1V1T2 / T1V2
-Substitution
P2 = (102)(4.52)(261) / (296)(4.83)
-Simplification
P2 = 120331.44 / 1429.68
-Result
P2 = 84.16 kPa
Answer:
C. Hold polar molecules together.
Explanation:
This is just something you have to know.
Answer:
Percent of gold = 75/100 × 100
= 75 %
Explanation:
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