KE = (1/2)mv^2
m = 10.0 kg
v = 5.00 m/a
KE = (1/2)(10.0)(5.00)^2 = (1/2)(10.0)(25.0) = 125 J
C. 125 J
Answer:
<em>56.4 m</em>
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Explanation:
volume increases by factor of 6, i.e
= 6
Initial temperature T1 at bottom of lake = 5.24°C = 278.24 K
Final temperature T2 at top of lake = 18.73°C = 291.73 K
NB to change temperature from °C to K we add 273
Final pressure P2 at the top of the lake = 0.973 atm
Initial pressure P1 at bottom of lake = ?
Using the equation of an ideal gas
= 
P1 =
= 
P1 = 5.57 atm
5.57 atm = 5.57 x 101325 = 564380.25 Pa
Density Ρ of lake = 1.02 g/
= 1020 kg/
acceleration due to gravity g = 9.81 
Pressure at lake bottom = pgd
where d is the depth of the lake
564380.25 = 1020 x 9.81 x d
d =
= <em>56.4 m</em>
Answer:
Formula and structure: The chemical formula of calcium chloride is CaCl2, and its molar mass is 110.983 g/mol. It is an ionic compound consisting of the calcium cation (Ca2+) and two chlorine anions (Cl-). The bivalent calcium metal forms an ionic bond with two chlorine atoms, as shown below.
2C₃H₇OH + 9O₂ = 6CO₂ + 8H₂O
V(O₂)=12.0 dm³
n(C₃H₇OH)=0.1 mol
n(O₂)=12.0 dm³/22.4 dm³/mol=0.5357 mol
C₃H₇OH : O₂ 2:9 1:4.5
0.1:0.5357
oxygen in excess
V(CO₂)=3Vm*n(C₃H₇OH)
V(CO₂)=3*22.4*0.1=6.72 dm³