Answer:
A. The rate of heat transfer through the material would increase.
Explanation:
To calculate the heat transfer in a heat exchanger you decide that there is not heat leakage to the surroundings, that means that magnitude of the two transfer rates will be equal. Any heat lost by the hot fluid, is gained by the cold fluid. The equation that describes this is Q = m×Cp×dT
Where:
heat = mass flow ×specific heat capacity × temperature difference
So if we increase the rate of flow of cooling water and the other variables that ypu can control remain the same, the result is that the rate of heat transfer through the material would increase, as it is stated in option a.
Answer:
The endpoint volume is 50.52 ± 0.14 mL
Explanation:
In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:
V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)
V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL
V = 50.52 ± 0.14 mL
It is necessary to consider the sum of the errors too.
75% i think it is consumers
25% i think it is producer
Answer:
The concentration the student should write down in her lab is 2.2 mol/L
Explanation:
Atomic mass of the elements are:
Na: 22.989 u
S: 32.065 u
O: 15.999 u
Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.
Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.
For mole(s) of Na2S2O3 = (mass taken)/(molar mass)
= (17.240 g)/(158.105 g/mol) = 0.1090 mole.
Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)
= 0.05029 L.
To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:
= (moles of sodium thiosulfate)/(volume of solution in L)
= (0.1090 mole)/(0.05029 L)
= 2.1674 mol/L
