The specific heat of metal is c = 3.433 J/g*⁰C.
<h3>Further explanation</h3>
Given
mass of metal = 68.6 g
t metal = 100 °C
mass water = 84 g
t water = 20 °C
final temperature = 52.1 °C
Required
The specific heat
Solution
Heat can be formulated :
Q = m.c.Δt
Q absorbed by water = Q released by metal
84 x 4.184 x (52.1-20)=68.6 x c x (100-52.1)
11281.738=3285.94 x c
c = 3.433 J/g*⁰C.
the mass in grams of one mole substance is the molar mass of the element
V=4.8 L
c=5.0 mol/L
M(Mg)=24.3 g/mol
1) n(HCl)=cv
2) m(Mg)=M(Mg)n(HCl)/2
3) m(Mg)=M(Mg)cv/2
m(Mg)=24.3*5*4.8/2=291.6 g
<u>Given:</u>
Moles of gas, n = 1.50 moles
Volume of cylinder, V = 15.0 L
Initial temperature, T1 = 100 C = (100 + 273)K = 373 K
Final temperature, T2 = 150 C = (150+273)K = 423 K
<u>To determine:</u>
The pressure ratio
<u>Explanation:</u>
Based on ideal gas law:
PV = nRT
P= pressure; V = volume; n = moles; R = gas constant and T = temperature
under constant n and V we have:
P/T = constant
(or) P1/P2 = T1/T2 ---------------Gay Lussac's law
where P1 and P2 are the initial and final pressures respectively
substituting for T1 and T2 we get:
P1/P2 = 373/423 = 0.882
Thus, the ratio of P2/P1 = 1.13
Ans: The pressure ratio is 1.13
