If you’re asking to balance the equation then:
Pb(NO3)2(aq) + 2KCl(aq) -> 2KNO3(aq) + PbCl2(s)
Just remember: the equations at the end is Cl not C12
Note: the small number on the bottom (subscripts) apply to the one element if it’s inside the bracket and if the small number is on the outside of the bracket it applies to all the elements. For example the 3 in (NO3)2 applied only to the O (oxygen) and the 2 applies to both N and O but don’t forget it’s multiplied. So it would be 2 N’s and 6 O’s bc the 3 multiplies with the 2 only for the O.
S- and P-Block elements are known as Representative elements.
Groups 1-2 and Groups 13-18
Answer:
The specific heat of gold is 0.129 J/g°C
Explanation:
Step 1: Data given
Mass of gold = 15.3 grams
Heat absorbed = 87.2 J
Initial temperature = 35.0 °C
Final temperature = 79.2 °C
Step 2:
Q = m*c*ΔT
⇒ Q =the heat absorbed = 87.2 J
⇒ m = the mass of gold = 15.3 grams
⇒ c = the specific heat of gold = TO BE DETERMINED
⇒ ΔT = The change in temperature = T2 - T1 = 79.2 - 35.0 = 44.2 °C
87.2 J = 15.3g * c * 44.2°C
c = 87.2 / (15.3 * 44.2)
c = 0.129 J/g°C
The specific heat of gold is 0.129 J/g°C
Answer:
30 cm³
Explanation:
Step 1: Given data
- Density of aluminum (ρ): 2.7 g/cm³
- Mass of aluminum (m): 81 g
- Volume occupied by aluminum (V): ?
Step 2: Calculate the volume occupied by aluminum
The density of aluminum is equal to its mass divided by its volume.
ρ = m/V
V = m/ρ
V = 81 g / 2.7 g/cm³
V = 30 cm³
Answer:
I guess you just answered a lot of questions
Explanation:
Thanks for the points btw :)