V
1
/T
1
=V
2
/T
2
(900.0 mL) / (300.0 K) = (x) / (405.0 K); x = 1215 mL.
Change the 900 to 800, and the 300 to 27, then change the 405 to 132. And solve
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
Sodium fluoride- to brush teeth
Citric acid- orange juice for breakfast
Sodium hydroxide- cleaning agent
The answer is metal. Metals are always named first in ionic compounds, like KNO3 for example. I hope this helps!
