Answer:
The heat capacity of the calorimeter is 5.11 J/g°C
Explanation:
Step 1: Given data
50.0 mL of water with temperature of 80.0 °C
Specific heat capacity of water = 4.184 J/g°C
Consider the density of water = 1g/mL
50.0 mL of water in a calorimeter at 20.0 °C
Final temperature = 47.0 °C
Step 2: Calculate specific heat capacity of the water in calorimeter
Q = Q(cal) + Q(water)
Q(cal) = mass * C(cal) * ΔT
Qwater = mass * Cwater * ΔT
Qcal = -Qwater
mass(cal) * C(cal) * ΔT(cal) = mass(water) * C(water) * ΔT(water)
50 grams * C(cal) * (47.0 - 20.0) =- 50grams * 4.184 J/g°C * (47-80)
1350 * C(cal) = 6903.6
C(cal) = 5.11 J/g°C
The heat capacity of the calorimeter is 5.11 J/g°C
Answer:
One to two years accurately represents the training period of an animal control officer.
Explanation:
Answer:
71g (assuming experimental data)
Explanation:
The balanced equation for this reaction:
+ 
→ 
+ 
Molar mass of H2SO4 = 98.1 g/mol
molar mass of NaOH = 40g/mol
Molar mass of Na2SO4 = 142.04g/mol
⇒ 1 mole or 98.1g of H2SO4 will yield 1 ole of NaSO4; alternately, 2 moles or 49 ×2 = 80g of NaOH produces 1 mole of NaSO4.
<em>Therefore, limiting reactant is NaOH.</em>
Assuming actual experiment is 20g of NaOH,
1 mole - 40g
x moles - 20g = 
= 0.5 moles
⇒1 mole of Na2SO4 - 142.04g
∴ 0.5 moles = 142.04 × 0.5
<u>= 71.02g</u>
The decay process that is demonstrated by the equation above would be a beta decay. This type of radioactive decay involves the emission of a beta particle. It happens when one of the protons is transformed in the other. The beta particle here is represented by <span>0 -1e.</span>
Answer:
This is Aluminium.
3 Valency electrons.
Explanation:
The atomic number is 13 and the electronic configuration is
1s22s22p63s23p1 . This is Aluminium.
It has 3 valency electrons. (3s23p1).
