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.
Answer: the concentration of [CO]= 0.0532M
Explanation:
From The equation of reaction
2H2(g) + CO(g) ⇌ CH3OH(g)
Applying Kc= [CH3OH]/[H2]^2[[CO]
[CH3OH]= 0.00487
[CO]= x-0.00487
[H2]=(0.032-0.00487)^2=0.0271
Substitute into formula
Kc=[CH3OH]/[H2]^2[[CO]
35= 0.00487/(0.0271)^2(x-0.00487)
Simplify
x-0.00487=0.189
x= 0.00487+0.189=0.193moles
[CO]= n/C= 0.193/3.63= 0.0532M
Answer: In a transverse wave, the particles of the medium move perpendicular to the wave's direction of travel. Transverse waves are characterized by peaks and valleys, called crests and troughs. In a longitudinal wave, the particles of the medium move parallel to the wave's direction of travel.
Explanation:
For Ar :
1 mol ------------ 22.4 L ( at STP )
7.6 mol ---------- x L
x = 7.6 * 22.4
x = 170.24 L
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For C2H3:
1 mol ------------ 22.4 ( at STP)
0.44 mol --------- y L
y = 0.44 * 22.4
y = 9.856 L
hope this helps !.
