What is the specific heat of a substance if a mass of 10.0 kg increases in temperature from 10.0°C to 70.0°C when 2,520 J of heat is applied? Use q equals m C subscript p Delta T..

Plug in what you know
-Heat (q) is 2520
-Delta T (change in temperature) is 70-10 which is 60
-Mass (M) is 10kg
-specific heat (c) is unknown.

2520 = (10)C(60)
Use algebra to solve for C.
Multiply C by 60 and 10.
2520 = 600C
Divide both sides by 600 to get C by itself.
C = 4.2 J/g

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.

Go ask spider man