Na2C2O4(aq) + CaCl2(aq) -----> 2NaCl(aq) + CaC2O4(s)
Here, CaC2O4(s) is a precipitate in the reaction as a result of precipitation reaction or double displacement reaction.
As we know that double displacement reaction two metal ions displaces each other from their salt solutions.
As we know that precipitation reaction is a reaction in which precipitate is formed.
Answer:
In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as
c=4.18Jg∘C
Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.
Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of 1 g of that substance by 1∘C.
In water's case, you need to provide 4.18 J of heat per gram of water to increase its temperature by 1∘C.
What if you wanted to increase the temperature of 1 g of water by 2∘C ?
This will account for increasing the temperature of the first gram of the sample by n∘C, of the the second gramby n∘C, of the third gram by n∘C, and so on until you reach m grams of water.
And there you have it. The equation that describes all this will thus be
q=m⋅c⋅ΔT , where
q - heat absorbed
m - the mass of the sample
c - the specific heat of the substance
ΔT - the change in temperature, defined as final temperature minus initial temperature
In your case, you will have
q=100.0g⋅4.18Jg∘C⋅(50.0−25.0)∘C
q=10,450 J
0.004382166 Make sure to round to the right amount of Sig Figs
Answer:
20.9%
Explanation:
- The percentage by mass of solution is given by dividing the mass of solute in grams by the mass of solution in grams then multiplying it by 100%.
% Mass of solution = mass of solute/mass of solution × 100%
= (27.0 g/ 129.0 g) × 100%
= 20.93%
= 20.9%
Boiling water, steam from a cup of tea and ice melting