Answer:
If 1,079.75 Joules of heat are added to 77.75 grams of water, by 3.32 degrees Celsius the temperature of water will increase

Explanation:

Here , q = heat added / removed from the substance
m = mass of the substance taken
= Change in temperature
C = specific heat capacity of the substance
In liquid state the value of C for water is :

Given values :
q = 1079.75 J
m = 77.75 gram
Insert the value of C, m , q in the given equation
on transposing ,



Answer:

Explanation:
Hello,
In this case, we can consider that the given heat of combustion is indeed the heat of reaction since it corresponds to the combustion of propane, which is computed by using the heat formation of all the involved species as shown below:

Thus, since the heat of formation of gaseous carbon dioxide is -393.5 kJ/mol, water -241.8 kJ/mol and oxygen 0 kJ/mol, the heat of formation of propane is:

Best regards.
Answer:
The correct answer will be "4.60 g".
Explanation:
The given values are:
Volume of Butane = 7.96 mL
Density = 0.579 g/mL
As we know,
⇒ 
On putting the estimated values, we get
⇒ 
⇒ 
Answer:
When the concentration of F- exceeds 0.0109 M, BaF2 will precipitate.
Explanation:
Ba²⁺(aq) + 2 F⁻(aq) <----> BaF₂(s)
When BaF₂ precipitates, the Ksp relation is given by
Ksp = [Ba²⁺] [F⁻]²
[Ba²⁺] = 0.0144 M
[F⁻] = ?
Ksp = (1.7 × 10⁻⁶)
1.7 × 10⁻⁶ = (0.0144) [F⁻]²
[F⁻]² = (1.7 × 10⁻⁶)/0.0144 = 0.0001180555
[F⁻] = √0.0001180555 = 0.01086 M = 0.0109 M
Hope this Helps!!!