Standard free energy (∆G°) of hydrolysis of creatine phosphate is (A) -–51.4 KJ/mol (B) –43.1 KJ/mol (C) –30.5 KJ/mol (D) –15.9 KJ/mol

Standard free energy (∆G°) of hydrolysis of creatine phosphate is (A) -–51.4 KJ/mol (B) –43.1 KJ/mol (C) –30.5 KJ/mol (D) –15.9 KJ/mol
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Standard free energy (∆G°) of hydrolysis of ADP to AMP + Pi is (A) –43.3 KJ/mol (B) –30.5 KJ/mol (C) –27.6 KJ/mol (D) –15.9 KJ/mol

Description : Standard free energy (∆G°) of hydrolysis of ADP to AMP + Pi is (A) –43.3 KJ/mol (B) –30.5 KJ/mol (C) –27.6 KJ/mol (D) –15.9 KJ/mol

Last Answer : Answer : C

Standard free energy (∆G°) of hydrolysis of phosphoenolpyruvate is (A) –61.9 KJ/mol (B) –43.1 KJ/mol (C) –14.2 KJ/mol (D) –9.2 KJ/mol

Description : Standard free energy (∆G°) of hydrolysis of phosphoenolpyruvate is (A) –61.9 KJ/mol (B) –43.1 KJ/mol (C) –14.2 KJ/mol (D) –9.2 KJ/mol

Last Answer : Answer : A

Standard free energy (∆G°) of hydrolysis of ATP to ADP + Pi is (A) –49.3 KJ/mol (B) –4.93 KJ/mol (C) –30.5 KJ/mol (D) –20.9 KJ/mol

Description : Standard free energy (∆G°) of hydrolysis of ATP to ADP + Pi is (A) –49.3 KJ/mol (B) –4.93 KJ/mol (C) –30.5 KJ/mol (D) –20.9 KJ/mol

Last Answer : Answer : C

Given that the standard free energy change (∆G°) for the hydrolysis of ATP is –7.3 K cal/mol and that for the hydrolysis of Glucose 6-phosphate is –3.3 Kcal/mol, the ∆G° for the phosphorylation of glucose is Glucose + ATP → Glucose 6– Phosphate + ADP. (A) –10.6 Kcal/mol (B) –7.3 Kcal/mol (C) –4.0 Kcal/mol (D) +4.0 Kcal/mol

Description : Given that the standard free energy change (∆G°) for the hydrolysis of ATP is -7.3 K cal/mol and that for the hydrolysis of Glucose 6-phosphate is -3.3 Kcal/mol, the ∆G° for the phosphorylation of glucose is Glucose + ATP → ... -10.6 Kcal/mol (B) -7.3 Kcal/mol (C) -4.0 Kcal/mol (D) +4.0 Kcal/mol

Last Answer : Answer : C

The standard free energy of hydrolysis of terminal phosphate group of ATP is (A) –7,300 cal/mol (B) –8,300 cal/mol (C) 10,000 cal/mol (D) +7,300 cal/mol

Description : The standard free energy of hydrolysis of terminal phosphate group of ATP is (A) –7,300 cal/mol (B) –8,300 cal/mol (C) 10,000 cal/mol (D) +7,300 cal/mol

Last Answer : A

Under standard conditions (A) The free energy change ∆G°, is equal to 0 (B) The standard free energy change ∆G, is equal to 0 (C) The free energy change, ∆G°, is equal to the standard free energy change, ∆G° (D) Keq is equal to 1

Description : Under standard conditions (A) The free energy change ∆G°, is equal to 0 (B) The standard free energy change ∆G, is equal to 0 (C) The free energy change, ∆G°, is equal to the standard free energy change, ∆G° (D) Keq is equal to 1

Last Answer : C

Given standard enthalpy of formation of `CO(-110 "KJ mol"^(-1))` and `CO_(2)(-394 "KJ mol"^(-1))`. The heat of combustion when one mole of graphite bu

Description : Given standard enthalpy of formation of `CO(-110 "KJ mol"^(-1))` and `CO_(2)(-394 "KJ mol"^(-1))`. The heat of combustion when one mole of graphite bu

Last Answer : Given standard enthalpy of formation of `CO(-110 "KJ mol"^(-1))` and `CO_(2)(-394 "KJ mol"^(-1))`. The heat ... B. `-284` KJ C. `-394` KJ D. `-504` KJ

If a reaction has an enthalpy of -54.32 kJ/mol and an entropy of -354.2 J/(K*mol), what is the Gibbs free Energy at 54.3(degrees c)?

Description : If a reaction has an enthalpy of -54.32 kJ/mol and an entropy of -354.2 J/(K*mol), what is the Gibbs free Energy at 54.3(degrees c)?

Last Answer : DeltaG = DeltaH - TDeltaS dG = -54.32 kJ/mol - (54'32+273)K(-354.2J/molK) NB Thevtemperature is quoted in Kelvin(K) and the Entropy must be converted to kJ by dividing by '1000'/ Hence dG = ... 115.94 kJ/mol dG = (+)61.61 kJ/mol Since dG is positive, the reaction is NOT thermodynamically feasible.

The temperature at which the reaction : `Ag_(2)O(s)rarr 2Ag(s)+1//2O_(2)(g)` is at equilibrium is …….., Given `Delta H=30.5 KJ mol^(-1)` and `Delta S

Description : The temperature at which the reaction : `Ag_(2)O(s)rarr 2Ag(s)+1//2O_(2)(g)` is at equilibrium is …….., Given `Delta H=30.5 KJ mol^(-1)` and `Delta S

Last Answer : The temperature at which the reaction : `Ag_(2)O(s)rarr 2Ag(s)+1//2O_(2)(g)` is at equilibrium is .., ... 362.12 K` C. `262.12 K` D. `562.12 K`

For the given reaction , `A(g) rarr 2B(g) , Delta_(r)H= 30 kJ//"mole" Delta_(r)S = 150 J//mol` at `300 K` If `C_(P,A) = 20 J//K mol` and `C_(P,B) = 20

Description : For the given reaction , `A(g) rarr 2B(g) , Delta_(r)H= 30 kJ//"mole" Delta_(r)S = 150 J//mol` at `300 K` If `C_(P,A) = 20 J//K mol` and `C_(P,B) = 20

Last Answer : For the given reaction , `A(g) rarr 2B(g) , Delta_(r)H= 30 kJ//"mole" Delta_(r)S = 150 J//mol` at ... G` is negative D. At `T = 200, Delta G` is zero

For the reaction `Ag_(2)O(s)rarr 2Ag(s)+1//2O_(2)(g)` the value of `Delta H=30.56 KJ mol^(_1)` and `Delta S = 66 JK^(-1)mol^(-1)`. The temperature at

Description : For the reaction `Ag_(2)O(s)rarr 2Ag(s)+1//2O_(2)(g)` the value of `Delta H=30.56 KJ mol^(_1)` and `Delta S = 66 JK^(-1)mol^(-1)`. The temperature at

Last Answer : For the reaction `Ag_(2)O(s)rarr 2Ag(s)+1//2O_(2)(g)` the value of `Delta H=30.56 KJ mol^(_1)` and `Delta ... :- A. 373 K B. 413 K C. 463 K D. 493 K

If the enthaply change for the transition of liquid water to steam is 30 KJ `"mol"^(-1)` at `27^(@)` C . The entropy change for the process would be

Description : If the enthaply change for the transition of liquid water to steam is 30 KJ `"mol"^(-1)` at `27^(@)` C . The entropy change for the process would be

Last Answer : If the enthaply change for the transition of liquid water to steam is 30 KJ `"mol"^(-1)` at `27^(@)` C . The ... K^(-1)` D. `100 J mol^(-1)K^(-1)`

The highest energy level is present amongst the following in (A) 1, 3-Biphosphoglycerate (B) Creatine phosphate (C) Carbamoyl phosphate (D) Phosphoenol pyruvate

Description : The highest energy level is present amongst the following in (A) 1, 3-Biphosphoglycerate (B) Creatine phosphate (C) Carbamoyl phosphate (D) Phosphoenol pyruvate

Last Answer : Answer : D

From the steam table, determine the average constant pressure specific heat (c) of steam at 10 kPa and45.8 ˚C  A.1.79 kJ/ kg-˚C  B.10.28 kJ/ kg-˚C  C.30.57 kJ/ kg-˚C  D. 100.1 kJ/ kg-˚C Formula: h = c T ∆ ∆ From the steam table At 47.7 ˚C h= 2588.1 kJ/ kg At 43.8 ˚C h= 2581.1 kJ/ kg

Description : From the steam table, determine the average constant pressure specific heat (c) of steam at 10 kPa and45.8 ˚C  A.1.79 kJ/ kg-˚C  B.10.28 kJ/ kg-˚C  C.30.57 kJ/ kg-˚C  D. 100.1 kJ/ kg-˚C Formula: h = c T ∆ ∆ From the steam table At 47.7 ˚C h= 2588.1 kJ/ kg At 43.8 ˚C h= 2581.1 kJ/ kg

Last Answer : 1.79 kJ/ kg-˚C

If ∆G°= –2.3RT log Keq, the free energy for the reaction will be A BC + 10moles 10moles 10moles (A) –4.6 RT (B) –2.3 RT (C) +2.3 RT (D) +4.6 RT

Description : If ∆G°= –2.3RT log Keq, the free energy for the reaction will be A BC + 10moles 10moles 10moles (A) –4.6 RT (B) –2.3 RT (C) +2.3 RT (D) +4.6 RT

Last Answer : Answer : C

What is the standard deviation of the following data set rounded to the nearest tenth 52.1 45.5 51 48.8 43.6?

Description : What is the standard deviation of the following data set rounded to the nearest tenth 52.1 45.5 51 48.8 43.6?

Last Answer : Feel Free to Answer

What is the major form of caloric storage in human body? (A) ATP (B) Glycogen (C) Creatine phosphate (D) triacylglycerol

Description : What is the major form of caloric storage in human body? (A) ATP (B) Glycogen (C) Creatine phosphate (D) triacylglycerol

Last Answer : Answer : C

Transmethylation of guanido acetic acid gives (A) Creatine phosphate (B) Creatinine (C) Choline (D) n-methyl nicotinamide

Description : Transmethylation of guanido acetic acid gives (A) Creatine phosphate (B) Creatinine (C) Choline (D) n-methyl nicotinamide

Last Answer : Answer : A

The value of enthalpy change `(DeltaH)` for the reaction `C_(2)H_(5)OH (l)+3O_(2) (g) rarr 2CO_(2) (g) +3H_(2)O (l)` at `27^(@)C` is `-1366.5 kJ mol^(

Description : The value of enthalpy change `(DeltaH)` for the reaction `C_(2)H_(5)OH (l)+3O_(2) (g) rarr 2CO_(2) (g) +3H_(2)O (l)` at `27^(@)C` is `-1366.5 kJ mol^(

Last Answer : The value of enthalpy change `(DeltaH)` for the reaction `C_(2)H_(5)OH (l)+3O_(2) (g) rarr 2CO_(2) (g) + ... 1369. kJ` C. `-1364.0 kJ` D. `-1361.5 kJ`

What shows that the formation of CO 2 releases 393.5 kJ mol?

Description : What shows that the formation of CO 2 releases 393.5 kJ mol?

Last Answer : C(s) + O2(g) + CO2(g) + 393.5 kJ

In adipose tissue, glycerol-3-phosphate required for the synthesis of triglycerides comes mainly from (A) Hydrolysis of pre-existing triglycerides (B) Hydrolysis of phospholipids (C) Dihydroxyacetone phosphate formed in glycolysis (D) Free glycerol

Description : In adipose tissue, glycerol-3-phosphate required for the synthesis of triglycerides comes mainly from (A) Hydrolysis of pre-existing triglycerides (B) Hydrolysis of phospholipids (C) Dihydroxyacetone phosphate formed in glycolysis (D) Free glycerol

Last Answer : Answer : C

The correct values of ionization enthalpies(in KJ `"mol"^(-1)`) of Si, P, Cl, and S respectively are: a)`786, 1012, 999, 1256` b)`1012, 786, 999, 1256

Description : The correct values of ionization enthalpies(in KJ `"mol"^(-1)`) of Si, P, Cl, and S respectively are: a)`786, 1012, 999, 1256` b)`1012, 786, 999, 1256

Last Answer : The correct values of ionization enthalpies(in KJ `"mol"^(-1)`) of Si, P, Cl, and S respectively are: ... . `786,1012,1256,999` D. `786,999,1012,1256`

An element has successive ionization enthalpies as 940 (first),2080,3090,4140,7030,7870,16000 and 19500 kJ `mol^(-1)`. To which group of the periodic

Description : An element has successive ionization enthalpies as 940 (first),2080,3090,4140,7030,7870,16000 and 19500 kJ `mol^(-1)`. To which group of the periodic

Last Answer : An element has successive ionization enthalpies as 940 (first),2080,3090,4140,7030,7870,16000 and 19500 kJ `mol^(- ... belong? A. 14 B. 15 C. 16 D. 17

The heat of combustion of carbon and monoxide are -394 and -285 KJ `mol^(-1)` respectively. The heat of formation of CO in KJ `mol^(-1)` is :-

Description : The heat of combustion of carbon and monoxide are -394 and -285 KJ `mol^(-1)` respectively. The heat of formation of CO in KJ `mol^(-1)` is :-

Last Answer : The heat of combustion of carbon and monoxide are -394 and -285 KJ `mol^(-1)` respectively. The heat of formation ... B. `-109` C. `+218` D. `-218`

The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :

Description : The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :

Last Answer : The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g) ... mol^(-1)` D. `-92.0 KJ mol^(-1)`

For `CaCO_(3)(s)rarr CaO(s)+CO_(2)(g)` at `977^(@)C, Delta H = 174` KJ/mol , then `Delta E` is :-

Description : For `CaCO_(3)(s)rarr CaO(s)+CO_(2)(g)` at `977^(@)C, Delta H = 174` KJ/mol , then `Delta E` is :-

Last Answer : For `CaCO_(3)(s)rarr CaO(s)+CO_(2)(g)` at `977^(@)C, Delta H = 174` KJ/mol , then `Delta E` is :- A. 160 KJ B. 163.6 KJ C. 186.4 KJ D. 180 KJ

The enthalpy of formation of `CO(g), CO_(2)(g),N_(2)O(g)` and `N_(2)O_(4)(g)` is `-110,-393,+81` and 10 kJ / mol respectively. For the reaction `N_(2)

Description : The enthalpy of formation of `CO(g), CO_(2)(g),N_(2)O(g)` and `N_(2)O_(4)(g)` is `-110,-393,+81` and 10 kJ / mol respectively. For the reaction `N_(2)

Last Answer : The enthalpy of formation of `CO(g), CO_(2)(g),N_(2)O(g)` and `N_(2)O_(4)(g)` is `-110,-393,+81` and 10 ... is A. `-212` B. `+212` C. `+778` D. `-778`

`DeltaH_"vaporisation"` (KJ/mol) are given for the hydrides of halogens in the following graph. The hydride HF will correspond to

Description : `DeltaH_"vaporisation"` (KJ/mol) are given for the hydrides of halogens in the following graph. The hydride HF will correspond to

Last Answer : `DeltaH_"vaporisation"` (KJ/mol) are given for the hydrides of halogens in the following graph. The hydride HF will ... . (B)Q C. (C)R D. (D)S

The value of `Delta_(0) " for " [Ti(H_(2)O)_(6)]^(3+)` is found to be 240 kJ `mol^(-1)` then predict the colour of the complex using the following tab

Description : The value of `Delta_(0) " for " [Ti(H_(2)O)_(6)]^(3+)` is found to be 240 kJ `mol^(-1)` then predict the colour of the complex using the following tab

Last Answer : The value of `Delta_(0) " for " [Ti(H_(2)O)_(6)]^(3+)` is found to be 240 kJ `mol^(-1)` then ... -34)J-sec,N_(A)=6xx10^(23)c=3xx10^(8) m//sec)`

Consider the following reactions at `1000^(@)C`. (A) `Zn_((s))+1/2 O_(2(g)) rarr ZnO_((s)), DeltaG^(@)=-360` kJ/mol (B) `C_("(gra)")+1/2 O_(2(g)) rarr

Description : Consider the following reactions at `1000^(@)C`. (A) `Zn_((s))+1/2 O_(2(g)) rarr ZnO_((s)), DeltaG^(@)=-360` kJ/mol (B) `C_("(gra)")+1/2 O_(2(g)) rarr

Last Answer : Consider the following reactions at `1000^(@)C`. (A) `Zn_((s))+1/2 O_(2(g)) rarr ZnO_((s)), DeltaG^( ... a) and (b) are true D. both (a) and (b) false

Ionization enthalpy of Li is 520 kJ mol^(-1) while that of F is 1681 kJ mol^(-1) . Explain.

Description : Ionization enthalpy of Li is 520 kJ mol^(-1) while that of F is 1681 kJ mol^(-1) . Explain.

Last Answer : Ionization enthalpy of Li is 520 kJ mol-1 while that of F is 1681 kJ mol-1. Explain.

At which temperature would a reaction with AH = -92 kJ/mol, AS = -0.199kJ/(mol·K) be spontaneous?

Description : At which temperature would a reaction with AH = -92 kJ/mol, AS = -0.199kJ/(mol·K) be spontaneous?

Last Answer : the answer is 400

A 15-year-old girl was seen by her family physician because of increasing lethargy. She had a recent history of the "flu". Biochemistry tests show that she has renal impairement. serum sodium 140 mmol/L (137 - 144) serum potassium 4.2 mmol/L (3.5 - 4.9) serum urea 28 mol/L (2.5 - 7.5) serum creatinine 280 µmol/L (60 - 110) Her condition does not improve after several weeks on corticosteroid therapy, so a renal biopsy is performed. The biopsy demonstrates the presence of segmental sclerosis of 3 of 10 glomeruli identified in the biopsy specimen. Immunofluorescence studies and electron microscopy do not reveal evidence for immune deposits. What is the most appropriate advice to give regarding her condition? 1) She has an underlying malignancy 2) She may require a renal transplant in 10 years

Description : A 15-year-old girl was seen by her family physician because of increasing lethargy. She had a recent history of the "flu". Biochemistry tests show that she has renal impairement. ... will likely develop a restrictive lung disease 5) She will probably improve with additional corticosteroid therapy

Last Answer : Answers-2 The findings point to focal segmental glomerulosclerosis (FSGS), which leads to chronic renal failure in half of cases. The lack of resolution with corticosteroid therapy and the progression to chronic renal failure is what sets FSGS apart from minimal change disease.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Last Answer : 22

An engine working on air standard Otto cycle has a cylinder diameter of 10 cm and stroke length of 15 cm. The ratio of specific heats for air is 1.4. If the clearance volume is 196.3 cc and the heat supplied per kg of air per cycle is 1800 kJ/kg, then work output per cycle per kg of air is a. 879.1 kJ/ b. 890.2 kJ c. 895.3 kJ d. 973.5 kJ

Description : An engine working on air standard Otto cycle has a cylinder diameter of 10 cm and stroke length of 15 cm. The ratio of specific heats for air is 1.4. If the clearance volume is 196.3 cc and the heat supplied per kg of air per ... kg of air is a. 879.1 kJ/ b. 890.2 kJ c. 895.3 kJ d. 973.5 kJ

Last Answer : ANSWER d. 973.5 kJ

Adipose tissue which is a store house for triacyl glycerol synthesis the same using (A) The glycerol released by hydrolysis of triacyl glycerol (B) The glycerol-3-phosphate obtained in the metabolism of glucose (C) 2-phosphoglycerate (D) 3-phosphoglycerate

Description : Adipose tissue which is a store house for triacyl glycerol synthesis the same using (A) The glycerol released by hydrolysis of triacyl glycerol (B) The glycerol-3-phosphate obtained in the metabolism of glucose (C) 2-phosphoglycerate (D) 3-phosphoglycerate

Last Answer : Answer : B

Synthesis of glutamine is accompanied by the hydrolysis of (A) ATP (B) ADP (C) TPP (D) Creatin phosphate

Description : Synthesis of glutamine is accompanied by the hydrolysis of (A) ATP (B) ADP (C) TPP (D) Creatin phosphate

Last Answer : Answer : A

The hydrolysis of Glucose-6-phosphate is catalysed by a specific phosphatase which is found only in (A) Liver, intestines and kidneys (B) Brain, spleen and adrenals (C) Striated muscle (D) Plasma

Description : The hydrolysis of Glucose-6-phosphate is catalysed by a specific phosphatase which is found only in (A) Liver, intestines and kidneys (B) Brain, spleen and adrenals (C) Striated muscle (D) Plasma

Last Answer : A

Given that standard heat enthalpy of `CH_(4), C_(2)H_(4)` and `C_(3)H_(8)` are -17.9, 12.5, -24.8 Kcal/mol. The `Delta H` for `CH_(4)+C_(2)Hrarr C_(3)

Description : Given that standard heat enthalpy of `CH_(4), C_(2)H_(4)` and `C_(3)H_(8)` are -17.9, 12.5, -24.8 Kcal/mol. The `Delta H` for `CH_(4)+C_(2)Hrarr C_(3)

Last Answer : Given that standard heat enthalpy of `CH_(4), C_(2)H_(4)` and `C_(3)H_(8)` are -17.9, 12.5, -24.8 Kcal ... . `-30.2` Kcal C. 55.2 Kcal D. `-19.4` Kcal

Find out the wrong number in the sequence. 52, 51, 48, 43, 34, 27, 16 (A) 27 (B) 34 (C) 43 (D) 48

Description : Find out the wrong number in the sequence. 52, 51, 48, 43, 34, 27, 16 (A) 27 (B) 34 (C) 43 (D) 48

Last Answer : Answer: B The series is decreese with order 1, 3, 5, 7, 9, 11. Therefore expected series will be 52, 51, 48, 43, 36, 27, 16. Hence 34 is wrong in seriese there must be 36.

Insert the missing number: 36/62, 39/63, 43/61, 48/64, ? (A) 51/65 (B) 56/60 (C) 54/60 (D) 33/60

Description : Insert the missing number: 36/62, 39/63, 43/61, 48/64, ? (A) 51/65 (B) 56/60 (C) 54/60 (D) 33/60

Last Answer : Answer: C Observer 36/62, 39/63, 43/61, 48/64, ? series Upper part is increase by 3, 4, 5, 6 i.e 36+3= 39, 39+4 = 43, 43+5 = 48, 48 + 6 = 54.

Two half-reactions of an electrochemical cell are given below : MnO–4 (aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I), E° = 1.51 V Sn2+ (aq) → Sn4+ (aq) + 2e–, E° = + 0.15 V. Construct the redox equation from the standard potential of the cell and predict if the reaction is reactant favoured or product favoured. -Chemistry

Description : Two half-reactions of an electrochemical cell are given below : MnO-4 (aq) + 8H+ (aq) + 5e- → Mn2+ (aq) + 4H2O (I), E° = 1.51 V Sn2+ (aq) → Sn4+ (aq) + 2e- ... the redox equation from the standard potential of the cell and predict if the reaction is reactant favoured or product favoured. -Chemistry

Last Answer : The reactions can be represented at anode and at cathode in the following ways : At anode (oxidation) : Sn2+ → = Sn4+ (aq) + 2e- ] 5 E° = + 0.15 V At cathode (reduction) : MnO-4(aq) + 8H+ ... = E°cathode - E°anode = 1.51 - 0.15 = + 1.36 V ∴ Positive value of E°cell favours formation of product.

Two half-reactions of an electrochemical cell are given below : MnO–4 (aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I), E° = 1.51 V Sn2+ (aq) → Sn4+ (aq) + 2e–, E° = + 0.15 V. Construct the redox reaction equation from the two half-reactions and calculate the cell potential from the standard potentials and predict if the reaction is reactant or product favoured. -Chemistry

Description : Two half-reactions of an electrochemical cell are given below : MnO-4 (aq) + 8H+ (aq) + 5e- → Mn2+ (aq) + 4H2O (I), E° = 1.51 V Sn2+ (aq) → Sn4+ (aq) + ... -reactions and calculate the cell potential from the standard potentials and predict if the reaction is reactant or product favoured. -Chemistry

Last Answer : The reactions can be represented at anode and at cathode in the following ways : At anode (oxidation) : Sn2+ → Sn4+ (aq) + 2e- ] 5 E° = + 0.15 V Af cathode (reduction) : MnO-4(aq) + 8H+ ( ... = E°cathode - E°anode = 1.51 - 0.15 = + 1.36 V ∴ Positive value of E°cell favours formation of product.

A gas is compressed in a cylinder by a movable piston to a volume onehalf its original volume. During the process 300 kJ heat left the gas and internal energy remained same. The work done on gas in Nm will be  (a) 300 Nm  (b) 300,000 Nm  (c) 30 Nm  (d) 3000 Nm  (e) 30,000 Nm.

Description : A gas is compressed in a cylinder by a movable piston to a volume onehalf its original volume. During the process 300 kJ heat left the gas and internal energy remained same. The work done on gas in Nm will be  (a) 300 Nm  (b) 300,000 Nm  (c) 30 Nm  (d) 3000 Nm  (e) 30,000 Nm.

Last Answer : Answer : b

Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Description : Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Last Answer : Here, the arranged data is 6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, 43 Total number of observations = 10 Here, 10 is an even number , therefore median will be the mean of (10 / 2)th and (10 / 2 + 1)th observation. ... ⇒ 3x + 12 / 2 = 24 ⇒ 3x - 12 = 48 ⇒ 3x = 60 ⇒ x = 20 ∴ The value of x = 20

Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Description : Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Last Answer : Here, the arranged data is 6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, 43 Total number of observations = 10 Here, 10 is an even number , therefore median will be the mean of (10 / 2)th and (10 / 2 + 1)th observation. ... ⇒ 3x + 12 / 2 = 24 ⇒ 3x - 12 = 48 ⇒ 3x = 60 ⇒ x = 20 ∴ The value of x = 20

A butt weld has been made by MMA in a 50mm thick impact tested steel plate. The welding electrode used was a basic type, 5mm diameter. The measured welding parameters for one of the runs was 250 amps, 24 volts, using DC+ve polarity, and a ROL:of 180mm per min. What would be the arc energy? a) Insufficient information given to calculate the arc energy b) 1.9 kJ/mm c) 1.87 kJ/mm d) 2.0 kJ/mm

Description : A butt weld has been made by MMA in a 50mm thick impact tested steel plate. The welding electrode used was a basic type, 5mm diameter. The measured welding parameters for one of the runs was 250 amps, 24 volts, using DC ... given to calculate the arc energy b) 1.9 kJ/mm c) 1.87 kJ/mm d) 2.0 kJ/mm

Last Answer : d) 2.0 kJ/mm

In 2000, a part of Rs. 30 lakhs was invested in Company P and the rest was invested in Company Q for one year. The total interest received was Rs. 2.43 lakhs. What was the amount invested in Company P? A. Rs.9 lakh B. Rs.11 lakh C.Rs. 12 lakh D.Rs.18 lakh E. None of these

Description : In 2000, a part of Rs. 30 lakhs was invested in Company P and the rest was invested in Company Q for one year. The total interest received was Rs. 2.43 lakhs. What was the amount invested in Company P? A. Rs.9 lakh B. Rs.11 lakh C.Rs. 12 lakh D.Rs.18 lakh E. None of these

Last Answer : D.Rs.18 lakh

A flywheel weighs 400 kg and has a radius of gyration of 30 cm about its axis of rotation. It is mounted on a shaft subjected to a moment of 40 kgm. If the flywheel starts from restits speed after 3 seconds wil be a.10.9 rad/sec b.19.0 rad/sec c.51.2 rad/sec d.3.33 rad/sec e.16.35 rad/sec

Description : A flywheel weighs 400 kg and has a radius of gyration of 30 cm about its axis of rotation. It is mounted on a shaft subjected to a moment of 40 kgm. If the flywheel starts from restits speed after 3 seconds wil be a.10.9 rad/sec b.19.0 rad/sec c.51.2 rad/sec d.3.33 rad/sec e.16.35 rad/sec

Last Answer : d. 3.33 rad/sec