Mass/volume = density
mass = (440 mg)*(1g)/(1000mg) = 0.440g
volume = (1000cm)(1000cm)(t)
where t = thickness
density = 2.70 g/cm^3 = (0.440g)/((1000cm)(1000cm)(t))
multiply both sides by 't' and divide both sides by (2.70g/cm^3)
t = (0.440) / ((1000cm)(1000cm)(2.70)) = 1.629x10^-7 cm
t = (1.629 x 10^-7 cm)*(1000000 micrometers)/(1 cm) = 0.1629 micrometers
Answer is t = 0.1629 micrometers
Answer:
823.7g
Explanation:
Using the formula as follows:
Q = m × c × ∆T
Where;
Q = amount of heat (J)
m = mass of substance (g)
c = specific heat capacity (J/g°C)
∆T = change in temperature (°C)
Using the information given in this question as follows:
Q = 6,400 J
m = ?
c of soil = 0.840 J/g°C
∆T = 9.25°C
Using Q = mc∆T
m = Q ÷ c∆T
m = 6,400 ÷ (0.840 × 9.25)
m = 6400 ÷ 7.77
m = 823.7g
Answer: Igneous rocks may be simply classified according to their chemical/mineral composition as felsic, intermediate, mafic, and ultramafic, and by texture or grain size: intrusive rocks are course grained (all crystals are visible to the naked eye) while extrusive rocks may be fine-grained (microscopic crystals) or glass.
Explanation: Hope this helped! :)
Answer: The rate constant is 
Explanation ;
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = age of sample = 4.26 min
a = initial amount of the reactant = 2.56 mg
a - x = amount left after decay process = 2.50 mg
Now put all the given values in above equation to calculate the rate constant ,we get
Thus rate constant is [tex]0.334s^{-1}
<span><span>LiF, LiCl, LiBr, LiI, LiAtNaF, NaCl, NaBr, NaI, NaAtKF, KCl, KBr, KI, KAt</span><span>RbF, RbCl, RbBr, RbI, RbAt CsF, CsCl, CsBr, CsI, CsAt FrF, FrCl, FrBr, FrI, FrAt<span>
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