100 mL of `PH_(3)` on decomposition produced phosphorus and hydrogen. The change in volume is

100 mL of `PH_(3)` on decomposition produced phosphorus and hydrogen. The change in volume is
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100 mL of `PH_(3)` on decomposition produced phosphorus and hydrogen. The change in volume is
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100 mL of `PH_(3) on decomposition produced phosphorus and hydrogen. The change in volume is :-

Description : 100 mL of `PH_(3) on decomposition produced phosphorus and hydrogen. The change in volume is :-

Last Answer : 100 mL of `PH_(3) on decomposition produced phosphorus and hydrogen. The change in volume is :- A. 50 ... . 500 mL decrease C. 900 mL decrease D. nil

The oxidation number of phosphorus in `PH_(4)^(+),PO_(2)^(3-),PO_(4)^(3-) and PO_(3)^(3-)` are respectively :-

Description : The oxidation number of phosphorus in `PH_(4)^(+),PO_(2)^(3-),PO_(4)^(3-) and PO_(3)^(3-)` are respectively :-

Last Answer : The oxidation number of phosphorus in `PH_(4)^(+),PO_(2)^(3-),PO_(4)^(3-) and PO_(3)^(3-)` are respectively :- A. ... . `+3,-3,+5,+1` D. `-3,+1,+5,+3`

The water produced during metabolic reactions in an adult is about (A) 100 ml/day (B) 300 ml/day (C) 500 ml/day (D) 700 ml/day

Description : The water produced during metabolic reactions in an adult is about (A) 100 ml/day (B) 300 ml/day (C) 500 ml/day (D) 700 ml/day

Last Answer : Answer : B

what is the density of CO GAS IF 0.196 g occupied a volume of 100 ml?

Description : what is the density of CO GAS IF 0.196 g occupied a volume of 100 ml?

Last Answer : 100ml = 1 dm3 0.196g = 196x10-6kg Density = mass/volume Density of CO = 196x10-6 / 1 = 196x10-6 kg/dm3

Urea clearance is the (A) Amount of urea excreted per minute (B) Amount of urea present in 100 ml of urine (C) Volume of blood cleared of urea in one minute (D) Amount of urea filtered by glomeruli in one minute

Description : Urea clearance is the (A) Amount of urea excreted per minute (B) Amount of urea present in 100 ml of urine (C) Volume of blood cleared of urea in one minute (D) Amount of urea filtered by glomeruli in one minute

Last Answer : Answer : A

The formula to calculate maximum urea clearance is U V× B , where U denotes (A) Concentration of urea in urine in gm/24hr (B) Concentration of urea in urine in mg/100 ml (C) Concentration of urea in blood in mg/100 ml (D) Volume of urine in ml/mt

Description : The formula to calculate maximum urea clearance is U V× B , where U denotes (A) Concentration of urea in urine in gm/24hr (B) Concentration of urea in urine in mg/100 ml (C) Concentration of urea in blood in mg/100 ml (D) Volume of urine in ml/mt

Last Answer : Answer : B

Which of the following statements are true of a patient with hyperglycemia and hyponatremia? A. The sodium concentration must be corrected by 5 mEq. per 100 mg. per 100 ml. elevation in blood glucose. B. With normal renal function, this patient is likely to be volume overloaded. C. Proper fluid therapy would be unlikely to include potassium administration. D. Insulin administration will increase the potassium content of cells. E. Early in treatment adequate urine output is a reliable measure of adequate volume resuscitation.

Description : Which of the following statements are true of a patient with hyperglycemia and hyponatremia? A. The sodium concentration must be corrected by 5 mEq. per 100 mg. per 100 ml. elevation in blood ... . E. Early in treatment adequate urine output is a reliable measure of adequate volume resuscitation.

Last Answer : Answer: D DISCUSSION: Each 100-mg. per 100 ml. elevation in blood glucose causes a fall in serum sodium concentration of approximately 2 mEq. per liter. Excess serum glucose ... correction of the patient's associated acidosis produce movement of potassium ions into the intracellular compartment

What would be the heart rate of a person if the cardiac output is 5 L, blood volume in the ventricles at the end of diastole is 100 mL and at the end of ventricular systole is 50 mL ? (a) 125 beats per minute (b) 50 beats per minute (c) 75 beats per minute (d) 100 beats per minute

Description : What would be the heart rate of a person if the cardiac output is 5 L, blood volume in the ventricles at the end of diastole is 100 mL and at the end of ventricular systole is 50 mL ? (a) 125 beats per minute (b) 50 beats per minute (c) 75 beats per minute (d) 100 beats per minute

Last Answer : (d) 100 beats per minute

The Volume of distribution of a drug administered at a dose of 300 mg and exhibiting 30 microgram/mL instantaneous concentration in plasma shall be (A)10 L (B) 100 L (C) 1.0 L (D) 0.10 L

Description : The Volume of distribution of a drug administered at a dose of 300 mg and exhibiting 30 microgram/mL instantaneous concentration in plasma shall be (A)10 L (B) 100 L (C) 1.0 L (D) 0.10 L

Last Answer : (A)10 L

If the total amount of a drug present in the body at a given moment is 2.0 g and its plasma concentration is 25 μg/ml, its volume of distribution is: A. 100 L B. 80 L C. 60 L D. 50 L

Description : If the total amount of a drug present in the body at a given moment is 2.0 g and its plasma concentration is 25 μg/ml, its volume of distribution is: A. 100 L B. 80 L C. 60 L D. 50 L

Last Answer : B. 80 L

2 If the total amount of a drug present in the body at a given moment is 2.0 g and its plasma concentration is 25 μg/ml, its volume of distribution is: A. 100 L B. 80 L C. 60 L D. 50 L

Description : 2 If the total amount of a drug present in the body at a given moment is 2.0 g and its plasma concentration is 25 μg/ml, its volume of distribution is: A. 100 L B. 80 L C. 60 L D. 50 L

Last Answer : B. 80 L

If the total amount of a drug present in the body at a given moment is 2.0 g and its plasma concentration is 25 μ g/ml, its volume of distribution is: A. 100 L B. 80 L C. 60 L D. 50 L

Description : If the total amount of a drug present in the body at a given moment is 2.0 g and its plasma concentration is 25 μ g/ml, its volume of distribution is: A. 100 L B. 80 L C. 60 L D. 50 L

Last Answer : B. 80 L

Liquid "X" is at equilibrium with its vapor in a cylinder and piston apparatus. When the volume of the space above the liquid is 50 ml and the temperature 25!, the vapor pressure of "X" is 120 torr. What will the vapor pressure of "X" be when the volume above the liquid is 100 ml and the temperature 25!C? Some liquid is always present. w) 30 torr x) 60 torr y) 120 torr z) 240 torr

Description : Liquid "X" is at equilibrium with its vapor in a cylinder and piston apparatus. When the volume of the space above the liquid is 50 ml and the temperature 25!, the vapor pressure of "X" is 120 torr. What will the ... 25!C? Some liquid is always present. w) 30 torr x) 60 torr y) 120 torr z) 240 torr

Last Answer : ANSWER: Y -- 120 TORR

Arrange the following compounds in decreasing order of reactivity for EAR :- (I) `CH_(2)=CH_(2)` (II) `CH_(3)-CH=CH_(2)` (III) `Ph_(2)C=CH-CH_(3)` (IV

Description : Arrange the following compounds in decreasing order of reactivity for EAR :- (I) `CH_(2)=CH_(2)` (II) `CH_(3)-CH=CH_(2)` (III) `Ph_(2)C=CH-CH_(3)` (IV

Last Answer : Arrange the following compounds in decreasing order of reactivity for EAR :- (I) `CH_(2)=CH_(2)` (II) `CH_(3)- ... I gt IV` D. `II gt III gt IV gt I`

White phosphorous on reaction with `NaOH` gives `PH_(3)` and

Description : White phosphorous on reaction with `NaOH` gives `PH_(3)` and

Last Answer : White phosphorous on reaction with `NaOH` gives `PH_(3)` and A. `Na_(2)HPO_(3)` B. `NaH_(2)PO_(2)` C. `NaH_(2)PO_(3)` D. `Na_(3)PO_(4)`

`P_(4)(s) + 3OH^(-) (aq) + 3H_(2)O (I) rarr PH_(3) (g) + 3H_(2)PO_(2)^(-) (aq)` In the above equation, the species getting oxidized and reduced respec

Description : `P_(4)(s) + 3OH^(-) (aq) + 3H_(2)O (I) rarr PH_(3) (g) + 3H_(2)PO_(2)^(-) (aq)` In the above equation, the species getting oxidized and reduced respec

Last Answer : `P_(4)(s) + 3OH^(-) (aq) + 3H_(2)O (I) rarr PH_(3) (g) + 3H_(2)PO_(2)^(-) (aq)` In the above equation ... ` C. `P_(4) and H_(2)O` D. `P_(4) and P_(4)`

(i) `NH_(3)` (ii) `N_(2)H_(4)` (iii) `HN_(3)` (iv) `PH_(3)` (v) `H_(2)S` (vi) `AsH_(3)` (vii) `SbH_(3)` (viii) `H_(2)Se` (ix) `H_(2)Te` Number of mole

Description : (i) `NH_(3)` (ii) `N_(2)H_(4)` (iii) `HN_(3)` (iv) `PH_(3)` (v) `H_(2)S` (vi) `AsH_(3)` (vii) `SbH_(3)` (viii) `H_(2)Se` (ix) `H_(2)Te` Number of mole

Last Answer : (i) `NH_(3)` (ii) `N_(2)H_(4)` (iii) `HN_(3)` (iv) `PH_(3)` (v) `H_(2) ... lone pair of electrons on the central atom is present in pure s-orbital

`PH_(3) ("anhydrous") + HBr ("anhydrous") rarr X`. Identify X ?

Description : `PH_(3) ("anhydrous") + HBr ("anhydrous") rarr X`. Identify X ?

Last Answer : `PH_(3) ("anhydrous") + HBr ("anhydrous") rarr X`. Identify X ? A. `H_(3)BrO_(3)` B. `PH_(4)Br` C. `Br_(2)` D. `P_(4)`

White phosphorous on reaction with `NaOH` gives `PH_(3)` and

Description : White phosphorous on reaction with `NaOH` gives `PH_(3)` and

Last Answer : White phosphorous on reaction with `NaOH` gives `PH_(3)` and A. `Na_(2)HPO_(3)` B. `NaH_(2)PO_(2)` C. `NaH_(2)PO_(3)` D. `Na_(3)PO_(4)`

`P_(4)(s) + 3OH^(-) (aq) + 3H_(2)O (I) rarr PH_(3) (g) + 3H_(2)PO_(2)^(-) (aq)` In the above equation, the species getting oxidized and reduced respec

Description : `P_(4)(s) + 3OH^(-) (aq) + 3H_(2)O (I) rarr PH_(3) (g) + 3H_(2)PO_(2)^(-) (aq)` In the above equation, the species getting oxidized and reduced respec

Last Answer : `P_(4)(s) + 3OH^(-) (aq) + 3H_(2)O (I) rarr PH_(3) (g) + 3H_(2)PO_(2)^(-) (aq)` In the above equation ... ` C. `P_(4) and H_(2)O` D. `P_(4) and P_(4)`

(i) `NH_(3)` (ii) `N_(2)H_(4)` (iii) `HN_(3)` (iv) `PH_(3)` (v) `H_(2)S` (vi) `AsH_(3)` (vii) `SbH_(3)` (viii) `H_(2)Se` (ix) `H_(2)Te` Number of mole

Description : (i) `NH_(3)` (ii) `N_(2)H_(4)` (iii) `HN_(3)` (iv) `PH_(3)` (v) `H_(2)S` (vi) `AsH_(3)` (vii) `SbH_(3)` (viii) `H_(2)Se` (ix) `H_(2)Te` Number of mole

Last Answer : (i) `NH_(3)` (ii) `N_(2)H_(4)` (iii) `HN_(3)` (iv) `PH_(3)` (v) `H_(2) ... lone pair of electrons on the central atom is present in pure s-orbital

`PH_(3) ("anhydrous") + HBr ("anhydrous") rarr X`. Identify X ?

Description : `PH_(3) ("anhydrous") + HBr ("anhydrous") rarr X`. Identify X ?

Last Answer : `PH_(3) ("anhydrous") + HBr ("anhydrous") rarr X`. Identify X ? A. `H_(3)BrO_(3)` B. `PH_(4)Br` C. `Br_(2)` D. `P_(4)`

The electronic spectrum of `[Ni(H_(2)O)_(6)]^(++)` shows as band at `8500cm^(-1)` due to d-d transiton . `[Ph_(4)As]_(2)[NiCl_(4)]` will have such a t

Description : The electronic spectrum of `[Ni(H_(2)O)_(6)]^(++)` shows as band at `8500cm^(-1)` due to d-d transiton . `[Ph_(4)As]_(2)[NiCl_(4)]` will have such a t

Last Answer : The electronic spectrum of `[Ni(H_(2)O)_(6)]^(++)` shows as band at `8500cm^(-1)` due to d-d transiton ... cm^(-1)` at A. 3778 B. 8500 C. 7250 D. 850

Among the following carbocations : (I) `Ph_(2)overset(+)"C"CH_(2)Me (II) PhCH_(2)CH_(2)overset(+)"C"HPh` `"(III) "Ph_(2)CHoverset(+)"C"HMe" (IV) "Ph_(

Description : Among the following carbocations : (I) `Ph_(2)overset(+)"C"CH_(2)Me (II) PhCH_(2)CH_(2)overset(+)"C"HPh` `"(III) "Ph_(2)CHoverset(+)"C"HMe" (IV) "Ph_(

Last Answer : Among the following carbocations : (I) `Ph_(2)overset(+)"C"CH_(2)Me (II) PhCH_(2)CH_( ... B. `IgtIIgtIIIgtIV` C. `IIgtIgtIVgtIII` D. `IgtIVgtIIIgtII`

Hydrogen bomb is based on the principle of - (1) Double decomposition (2) Artificial radioactivity (3) Nuclear fission (4) Nuclear fusion

Description : Hydrogen bomb is based on the principle of - (1) Double decomposition (2) Artificial radioactivity (3) Nuclear fission (4) Nuclear fusion

Last Answer : (4) Nuclear fusion Explanation: Hydrogen Bomb is based on the principle of Nuclear Fusion. Nuclear fusion is a reaction in which two or more atomic nuclei are combined to form one or more different atomic nuclei and subatomic particles.

Explain the disproportionation nature of decomposition of Hydrogen peroxide?

Description : Explain the disproportionation nature of decomposition of Hydrogen peroxide?

Last Answer : Ans. Decomposition of hydrogen peroxide produces water and oxygen. O.N Increases as well as decreases.

How many moles of oxygen are produced by the decomposition of six moles of patassium chlorate?

Description : How many moles of oxygen are produced by the decomposition of six moles of patassium chlorate?

Last Answer : Balanced equation: 2KClO3 --> 2KCl + 3O26 mol KClO3 = 3 x 2 mol KCl, so the number of moles O2 that can be produced by 6 mol KClO3 is:3 x 3 mol O2 = 9 mol O2.

How many moles of O2 are produced by the decomposition if 10 moles potassium chlorate 2KClO3- and 2KCl plus 3O2?

Description : How many moles of O2 are produced by the decomposition if 10 moles potassium chlorate 2KClO3- and 2KCl plus 3O2?

Last Answer : 15 moles of O2 are produced by the decomposition if 10 moles potassium chlorate 2 KClO3 and 2 KCl plus 3O2.

The gas produced in marshy places due to decomposition of vegetation is – (1) Carbon monoxide (2) Carbon dioxide (3) Sulphur dioxide (4) Methane

Description : The gas produced in marshy places due to decomposition of vegetation is – (1) Carbon monoxide (2) Carbon dioxide (3) Sulphur dioxide (4) Methane

Last Answer : (4) Methane Explanation: Methane gas is a hydrocarbon gas largely composed of methane formed when organic material or vegetation decays in the absence of air. Naturally occurring methane is mainly produced by the process of methanogenesis. It is also known as Swamp Gas.

The chemical decomposition produced by micro-organisms on certain organic matter is

Description : The chemical decomposition produced by micro-organisms on certain organic matter is

Last Answer : called fermentation.

Main component of sewage gas produced during anaerobic decomposition of organic waste (by suitable bacteria) during sewage disposal is (A) H2 (B) CH4 (C) CO2 (D) N2

Description : Main component of sewage gas produced during anaerobic decomposition of organic waste (by suitable bacteria) during sewage disposal is (A) H2 (B) CH4 (C) CO2 (D) N2

Last Answer : (B) CH4

Presence of carbon monoxide in atmosphere produced by decomposition of chlorophyll and haemoglobin breakdown of some animals, beyond TLV (>50 ppm) (A) Acts as a green house gas thereby raising earth's temperature (B) Causes asphyxia (C) Causes increase in sea level (D) Enhances the green house effect

Description : Presence of carbon monoxide in atmosphere produced by decomposition of chlorophyll and haemoglobin breakdown of some animals, beyond TLV (>50 ppm) (A) Acts as a green house gas thereby raising earth's ... (B) Causes asphyxia (C) Causes increase in sea level (D) Enhances the green house effect

Last Answer : (B) Causes asphyxia

The gas produced in marshy places due to decomposition of vegetation is (1) Carbon monoxide (2) Carbon dioxide (3) Sulphur dioxide (4) Methane

Description : The gas produced in marshy places due to decomposition of vegetation is (1) Carbon monoxide (2) Carbon dioxide (3) Sulphur dioxide (4) Methane

Last Answer : Methane

The decomposition of `N_(2)O_(5)` in chloroform was followed by measuring the volume of `O_(2)` gas evolved : `2N_(2)O_(5)(C C_(4))to2N_(2)O_(4)(C Cl_

Description : The decomposition of `N_(2)O_(5)` in chloroform was followed by measuring the volume of `O_(2)` gas evolved : `2N_(2)O_(5)(C C_(4))to2N_(2)O_(4)(C Cl_

Last Answer : The decomposition of `N_(2)O_(5)` in chloroform was followed by measuring the volume of `O_(2)` gas evolved : ... log"(90)/(100)` D. `(100)/(10xx500)`

The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorus acid is

Description : The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorus acid is

Last Answer : The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorus acid is A. zero B. two C. one D. three

The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorus acid is

Description : The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorus acid is

Last Answer : The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorus acid is A. zero B. two C. one D. three

(3) They are not easily combustible Explanation: Typical organic compounds are combustible by nature as they contain carbon. Organic compounds mostly consist of carbon, hydrogen, halogen, and phosphorus.

Description : (3) They are not easily combustible Explanation: Typical organic compounds are combustible by nature as they contain carbon. Organic compounds mostly consist of carbon, hydrogen, halogen, and phosphorus.

Last Answer : Which of the process is known as nitrification? (1) Reaction of Nitrogen Monoxide with oxygen to form nitric acid (2) Reaction of nitrogen dioxide with water to form nitric acid (3) Conversion of ammonia to nitrites (4) Conversion of nitrite to nitric oxide

Which of the following is used in making smoke bombs? (1) Sulphur (2) Phosphorus (3) Hydrogen (4) Carbon

Description : Which of the following is used in making smoke bombs? (1) Sulphur (2) Phosphorus (3) Hydrogen (4) Carbon

Last Answer : (4) Carbon Explanation: A smoke bomb is a firework designed to produce smoke upon ignition. Smoke bombs are useful to military units, airsoft games, paintball games, self-defense and pranks. The ... -style fireworks and later modified the formula to produce more smoke for a longer period of time.

Which of the following elements is not present in a nitrogenous base? (a) Hydrogen (b) Carbon (c) Phosphorus (d) Nitrogen

Description : Which of the following elements is not present in a nitrogenous base? (a) Hydrogen (b) Carbon (c) Phosphorus (d) Nitrogen

Last Answer : (c) Phosphorus

About 98 percent of the mass of every living organism is composed of just six elements including carbon, hydrogen, nitrogen, oxygen and (a) sulphur and magnesium (b) magnesium and sodium (c) calcium and phosphorus (d) phosphorus and sulphur.

Description : About 98 percent of the mass of every living organism is composed of just six elements including carbon, hydrogen, nitrogen, oxygen and (a) sulphur and magnesium (b) magnesium and sodium (c) calcium and phosphorus (d) phosphorus and sulphur.

Last Answer : (d) phosphorus and sulphur.

Which of the following is used in making smoke bombs? (1) Sulphur (2) Phosphorus (3) Hydrogen (4) Carbon

Description : Which of the following is used in making smoke bombs? (1) Sulphur (2) Phosphorus (3) Hydrogen (4) Carbon

Last Answer : Carbon

What is produced when phosphorus pentachloride is heated ?

Description : What is produced when phosphorus pentachloride is heated ?

Last Answer : When phosphorus pentachloride is heated, phosphorus trichloride and chlorine are produced.

Pick up the correct statement from the following: (A) Steel produced by open hearth process is milder than that obtained by the Bessemer process (B) Engineers prefer open hearth steel for structural purpose as it is more homogenous (C) Basic Bessemer process is suitable for converting poor ore containing a large proportion of sulphur and phosphorus into steel

Description : Pick up the correct statement from the following: (A) Steel produced by open hearth process is milder than that obtained by the Bessemer process (B) Engineers prefer open hearth steel ... is suitable for converting poor ore containing a large proportion of sulphur and phosphorus into steel

Last Answer : Answer: Option D

Hydrogen peroxide solution `(20 mL)` reacts quantitatively with a solution of `KMnO_(4) (20 mL)` acidified with dilute of `H_(2)SO_(4)`. The same volu

Description : Hydrogen peroxide solution `(20 mL)` reacts quantitatively with a solution of `KMnO_(4) (20 mL)` acidified with dilute of `H_(2)SO_(4)`. The same volu

Last Answer : Hydrogen peroxide solution `(20 mL)` reacts quantitatively with a solution of `KMnO_(4) (20 mL)` ... calculate the molarity of `H_(2)O_(2)`.

Why is the weight of 5 ml vinegar almost equal to its volume?

Description : Why is the weight of 5 ml vinegar almost equal to its volume?

Last Answer : Water has a density of 1 gram per cubic centimetre because it is defined that way and vinegar is mostly water.

How much volume (mL) is in a sip?

Description : How much volume (mL) is in a sip?

Last Answer : Uh. This is an educated guess, but I’m going to say a small sip may be between 5 and 10 mL, and a bigger sip (a drink) would be more like 10–15 mL.

The pressure is changed from 500 kPa to 250 kPa. What would you expect the new volume to be if the initial volume is 200 mL?

Description : The pressure is changed from 500 kPa to 250 kPa. What would you expect the new volume to be if the initial volume is 200 mL?

Last Answer : 400 mL(Explanation): again boyle's law PV=PV. Since the pressure is halved (500 to 250), then the volume must be doubled in order to maintain this equation. 200 x2=400 so that's the answer

Match column I with column II and select the correct option from the given codes. `{:(,"Column I",,"Column II"),(A.,"Tidal volume",(i),2500-3000 mL "o

Description : Match column I with column II and select the correct option from the given codes. `{:(,"Column I",,"Column II"),(A.,"Tidal volume",(i),2500-3000 mL "o

Last Answer : Match column I with column II and select the correct option from the given codes. `{:(,"Column I",,"Column II"),(A. ... (A,B,C,D,E),(iv,iii,ii,i,v):}`

The volume (in ml) of 0.5 M NaOH required for the complete reaction with 150 ml of 1.5M `H_(3)PO_(3)` solution is

Description : The volume (in ml) of 0.5 M NaOH required for the complete reaction with 150 ml of 1.5M `H_(3)PO_(3)` solution is

Last Answer : The volume (in ml) of 0.5 M NaOH required for the complete reaction with 150 ml of 1.5M `H_(3)PO_(3)` solution is A. 1350 B. 900 C. 1250 D. 1150