Stoichiometry The unit of dynamic viscosity is Both B & C Stoke Poise Gm/cm sec Both B & C Stoke Poise Gm/cm sec ANSWER DOWNLOAD EXAMIANS APP
Stoichiometry Heat of neutralisation of a strong acid and strong base is always a constant value, i.e., 57 KJ/Kg mole. This is because The strong base and strong acid reacts completely The salt formed does not hydrolyse The strong base and strong acid reacts in aqueous solution Only OH⁻ and H⁺ ions react in every case The strong base and strong acid reacts completely The salt formed does not hydrolyse The strong base and strong acid reacts in aqueous solution Only OH⁻ and H⁺ ions react in every case ANSWER DOWNLOAD EXAMIANS APP
Stoichiometry The latent heat of vaporisation Decreases with increased temperature Decreases as pressure increases All of these Becomes zero at the critical point Decreases with increased temperature Decreases as pressure increases All of these Becomes zero at the critical point ANSWER DOWNLOAD EXAMIANS APP
Stoichiometry A butane isomerisation process produces 70 kmole/hr of pure isobutane. A purge stream removed continuously, contains 85% n-butane and 15% impurity (mole%). The feed stream is n-butane containing 1% impurity (mole%). The flow rate of the purge stream will be 6 kmole/hr 5 kmole/hr 3 kmole/hr 4 kmole/hr 6 kmole/hr 5 kmole/hr 3 kmole/hr 4 kmole/hr ANSWER DOWNLOAD EXAMIANS APP
Stoichiometry An equation for calculating vapour pressure is given by, log10 P = A - B(t + c). Kopp's rule Trouton's rule Antonic equation Kistyakowsky equation Kopp's rule Trouton's rule Antonic equation Kistyakowsky equation ANSWER DOWNLOAD EXAMIANS APP
Stoichiometry Pick out the wrong unit conversion of temperature. °F = (°C + 17.778) x 1.8 Temperature difference of 1°K = 1°C = 9/5 °F °R = 273 + °F °C = (F- 32) x 0.555 °F = (°C + 17.778) x 1.8 Temperature difference of 1°K = 1°C = 9/5 °F °R = 273 + °F °C = (F- 32) x 0.555 ANSWER DOWNLOAD EXAMIANS APP
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