Ion knoe Wdym by “be able to describ’ so ima put it in my own words idr lol:)
if you talm bout some kentic energy or sum ok but other Dan dat ion knoe tbh
I can explain how transferring kinetic energy in and out of a substance can cause a change
Hello!
To solve this problem we're going to use the
Charles' Law. This Law describes the relationship between
Volume and Temperature in an ideal gas. Applying this law we have the following equation:

So, the final temperature is
54,23 °CHave a nice day!
21.4 g Al * (1 mol / 26.98 g ) * (2 mol Fe / 2 mol Al) = 0.793 mol Fe
91.3 g Fe2O3 * (1 mol / 159.69 g) * (2 mol Fe / 1 mol Fe2O3) = 1.14 mol Fe
0.793 mol Fe * (55.85 g / 1 mol) = 44.3 g Fe produced.
Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Hello:
In this case, we will use the Clapeyron equation:
P = ?
n = 8 moles
T = 250 K
R = 0.082 atm.L/mol.K
V = 6 L
Therefore:
P * V = n * R * T
P * 6 = 8 * 0.082* 250
P* 6 = 164
P = 164 / 6
P = 27.33 atm
Hope that helps!