Answer:
-10778.95 J heat must be removed in order to form the ice at 15 °C.
Explanation:
Given data:
mass of steam = 25 g
Initial temperature = 118 °C
Final temperature = 15 °C
Heat released = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 15 °C - 118 °C
ΔT = -103 °C
now we will put the values in formula
q = m . c . ΔT
q = 25 g × 4.186 J/g.°C × -103 °C
q = -10778.95 J
so, -10778.95 J heat must be removed in order to form the ice at 15 °C.
Answer:
B.) 1.3 atm
Explanation:
To find the new pressure, you need to use Gay-Lussac's Law:
P₁ / T₁ = P₂ / T₂
In this equation, "P₁" and "T₁" represent the initial pressure and temperature. "P₂" and "T₂" represent the final pressure and temperature. After converting the temperatures from Celsius to Kelvin, you can plug the given values into the equation and simplify to find P₂.
P₁ = 1.2 atm P₂ = ? atm
T₁ = 20 °C + 273 = 293 K T₂ = 35 °C + 273 = 308 K
P₁ / T₁ = P₂ / T₂ <----- Gay-Lussac's Law
(1.2 atm) / (293 K) = P₂ / (308 K) <----- Insert values
0.0041 = P₂ / (308 K) <----- Simplify left side
1.3 = P₂ <----- Multiply both sides by 308
Answer: The number of grams of 
in 1620 mL is 1.44 g
Explanation:
According to ideal gas equation:
P = pressure of gas = 1 atm (at STP)
V = Volume of gas = 1620 ml = 1.62 L (1L=1000ml)
n = number of moles = ?
R = gas constant =
T =temperature =
Mass of hydrogen =
The number of grams of in 1620 mL is 1.44 g
So, I don't know the answer for b but I think I found the answer for a.
I sincerely hope this isn't wrong
