Answer:
1.2029 J/g.°C
Explanation:
Given data:
Specific heat capacity of titanium = 0.523 J/g.°C
Specific heat capacity of 2.3 gram of titanium = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
1 g of titanium have 0.523 J/g.°C specific heat capacity
2.3 × 0.523 J/g.°C
1.2029 J/g.°C
Answer : The molar concentration of sucrose in the tea is, 0.0549 M
Explanation : Given,
Mass of sucrose = 3.765 g
Volume of solution = 0.200 L
Molar mass of sucrose = 342.3 g/mole
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :

Now put all the given values in this formula, we get:

Therefore, the molar concentration of sucrose in the tea is, 0.0549 M
Answer:
The answer to your question is 32.44 moles
Explanation:
Data
moles of Na₂CO₃ = ?
volume = 9.54 l
concentration = 3.4 M
Formula
Molarity = 
Solve for number of moles
number of moles = Molarity x volume
Substitution
Number of moles = (3.4)( 9.54)
Simplification
Number of moles = 32.44
Ans is B.Bromine. Only two elements are liquid at room temperature ( mercury and Bromine) . Helium and chlorine, arsenic are gases.