Given:
The heat content of the pizza is Q = 500 kcal
The volume of the cold water is V = 50 L
The mass of a liter of cold water is 1 kg.
Calculate the mass, M, of cold water.
M = (50 L)*(1 kg/L) = 50 kg
The specific heat of water is
c = 4.184 kJ/(kg-K)
Also,
1 kcal = 4184 J, therefore
Q = (500 kcal)*(4184 J/kcal) = 2.092 x 10⁶ J
Let ΔT = increase in temperature of the cold water, °C (same as K).
Then
Q = M*c*ΔT
or
ΔT = Q/(M*c)
Answer: The temperature increases by 10 °C (or 10 K)
Answer:
When the solutions (including inside and outside) were replaced by NaCl, the membrane potential changes from a negative value of -60 mV to a positive value of + 60 mV.
Explanation:
According to the Nernst equation the potential of the membrane for a two-compartment model of a cell for positive gradient of K⁺ ions is V = - 60 mVlogK'/K where K' = inside concentration of K⁺ ions and K = outside concentration of K⁺ ions. For a 10 fold excess of KCl in the inside compartment, K'/K = 10. So,
V = - 60 mVlogK'/K = - 60 mVlog10 = -60 mV.
For a negative gradient of Na ions is V = + 60 mVlogK'/K where K' = inside concentration of Na⁺ ions and K = outside concentration of Na⁺ ions. (Since the cell is selectively permeable to Na⁺ ions. So, Na ions to not flow out but in.)For a 10 fold excess of NaCl in the inside compartment, K'/K = 10. So,
V = + 60 mVlogK'/K = + 60 mVlog10 = +60 mV.
So, when the solutions (including inside and outside) were replaced by NaCl, the membrane potential changes from a negative value of -60 mV to a positive value of + 60 mV.
Gamma-Rays have the shortest wavelength.
Transparent solute cannot be separated from solvent by filtration or by passing through semipermeable membrane.
<h3>Can suspensions be separated by semipermeable membrane?</h3>
Suspensions are heterogeneous that are various from solutions or colloids.
The particles of a suspension are so large that they can often be seen with the eye. They are trapped by filters and semipermeable membranes.
Thus, Transparent solute cannot be separated from solvent by filtration.
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<span>Polydactyly in its most common form, is an autosomal dominant mutation. Let's represent the mutation as P and the normal state as +. Thus, if both parents are heterozygous (P+), both would have the mutation (and thus have six fingered). They have 3/4 chances of having a child with the extra digit, 1/4 normal. </span>
