Energy absorbed by Iron block E (iron) = 460.5 J
Energy absorbed by Copper block E (Copper) = 376.8 J
<u>Explanation:</u>
To find the heat absorbed, we can use the formula as,
q = m c ΔT
Here, Mass = m = 10 g = 0.01 kg
ΔT = change in temperature = 400 - 300 = 100 K = 100 - 273 = -173 °C
c = specific heat capacity
c for iron = 460.5 J/kg K
c for copper = 376.8 J/kg K
Plugin the values in the above equation, we will get,
q (iron) = 0.01 kg × 460.5 J/kg K × 100 K
= 460.5 J
q (copper) = 0.01 kg × 376.8 J/kg K × 100 K
= 376.8 J
The amount of HCl required for one experiment - 13.5 µl
the volume in terms of L - 13.5 x 10⁻⁶ L
the volume of HCl available - 0.250 L
since one experiment uses up - 13.5 x 10⁻⁶ L
then number of experiments - 0.250 L / 13.5 x 10⁻⁶ L = 1.8 x 10⁴ times
the experiment can be carried out 18000 times
Answer:
Increases I had this question
Answer:
[KMnO₄] = 0.71 M
Explanation:
Solute: Potassium permanganate, molar mass 158.04 g/mol
Solvent: Water, volume of 100 mL
We convert the mass of solute to moles:
11.3 g . 1mol / 158.04g = 0.071 moles
We asume the solvent's volume as the solution's volume
We convert the volume from mL to L → 100 mL . 1L / 1000 mL = 0.1L
Molarity (mol/L) → moles of solute in 1L of solution
Molarity → 0.071 moles / 0.1 L = 0.71M
