`P_(680)` is related with

`P_(680)` is related with
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`P_(680)` is related with A. Light B. Dark C. Both D. None
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`P_(680)` is calld PS II ! C-4 plants hav photorespiration.

Description : `P_(680)` is calld PS II ! C-4 plants hav photorespiration.

Last Answer : `P_(680)` is calld PS II ! C-4 plants hav photorespiration. A. If both As and R are correct and R is ... , R is wrong D. If both As and R are wrong.

What is bigger 0.680 or 0.079?

Description : What is bigger 0.680 or 0.079?

Last Answer : 0.0680 is bigger

What is 680÷20?

Description : What is 680÷20?

Last Answer : 34

In PS I the reaction centre chlorophyll a has an absorption peak at 700 nm, hence is called (A)____ while in PS II it has absorption maxima at 680 nm

Description : In PS I the reaction centre chlorophyll a has an absorption peak at 700 nm, hence is called (A)____ while in PS II it has absorption maxima at 680 nm

Last Answer : In PS I the reaction centre chlorophyll a has an absorption peak at 700 nm, hence is called (A) ... absorption maxima at 680 nm and is called (B)____

What is 4.680 rounded to the nearest tenth?

Description : What is 4.680 rounded to the nearest tenth?

Last Answer : 4.7

What is the answered to 0.04 divided by 680?

Description : What is the answered to 0.04 divided by 680?

Last Answer : 0.0001

Of the 40 randomly chosen students surveyed, 27 are involved in extracurricular activities at school. There are 680 students in the school. Predict the number of students in the school who are involved in extracurricular activities?

Description : Of the 40 randomly chosen students surveyed, 27 are involved in extracurricular activities at school. There are 680 students in the school. Predict the number of students in the school who are involved in extracurricular activities?

Last Answer : Jakd

A normal CD-ROM usually can store up to data? A) 680 KB B) 680 Bytes C) 680 MB D) 680 GB

Description : A normal CD-ROM usually can store up to data? A) 680 KB B) 680 Bytes C) 680 MB D) 680 GB

Last Answer : Answer : C

A 1000W carrier is amplitude modulated to a depth of 85%. Calculate the total power of the modulated wave? A. 1361.25 W B. 680.62 W C. 1425 W D. 712.5 W

Description : A 1000W carrier is amplitude modulated to a depth of 85%. Calculate the total power of the modulated wave? A. 1361.25 W B. 680.62 W C. 1425 W D. 712.5 W

Last Answer : A. 1361.25 W

The wavelength of light absorbed by Pr form of phytochrome is (a) 680 nm (b) 720 nm (c) 620 nm (d) 640 nm

Description : The wavelength of light absorbed by Pr form of phytochrome is (a) 680 nm (b) 720 nm (c) 620 nm (d) 640 nm

Last Answer : (a) 680 nm

One unit of electricity is equivalent to kcal heat units. a) 800 b) 860 c) 400 d) 680

Description : One unit of electricity is equivalent to kcal heat units. a) 800 b) 860 c) 400 d) 680

Last Answer : 860

32. One unit of electricity is equivalent to kcal heat units. a) 800 b) 860 c) 400 d) 680

Description : One unit of electricity is equivalent to kcal heat units. a) 800 b) 860 c) 400 d) 680

Last Answer : 860

One unit of electricity is equivalent to kcal heat units. c) 800 c) 400 d)860 d) 680

Description : One unit of electricity is equivalent to kcal heat units. c) 800 c) 400 d)860 d) 680

Last Answer : 860

114. One unit of electricity is equivalent to kcal heat units. a) 800 c) 400 b) 860 d) 680

Description : 114. One unit of electricity is equivalent to kcal heat units. a) 800 c) 400 b) 860 d) 680

Last Answer : 860

Modern lithium ion batteries provide average range of • 120-280 kilometres • 320-480 kilometres • 520-680 kilometres • 720-980 kilometres

Description : Modern lithium ion batteries provide average range of • 120-280 kilometres • 320-480 kilometres • 520-680 kilometres • 720-980 kilometres

Last Answer : • 320-480 kilometres

A rectangular grassy plot 110 m by 65 m has a gravel path 2.5 m wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq. Meter 1).Rs.680 2).Rs.860 3).Rs.608 4).None.

Description : A rectangular grassy plot 110 m by 65 m has a gravel path 2.5 m wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq. Meter 1).Rs.680 2).Rs.860 3).Rs.608 4).None.

Last Answer : 1).Rs.680 Exp: overall area of the grassy plot is 110 x 65 =7150 sq. meters.The area of the grassy plot after applying the gravel path is 105 x 60=6300 sq. meters.Area of the gravel path is 7150 – 6300 = 850 sq.mtrs.Hence x 0.80=Rs.680 .

A taxi driver make a trip from the plains to ooty which are 340km apart at an average speed of 60km/hr. In the return trip, he covers the speed distance at an average of 30 km/hr. the average speed of the taxi over the entire distance of 680 km is a) 30 b) 40 c) 50 d) 60

Description : A taxi driver make a trip from the plains to ooty which are 340km apart at an average speed of 60km/hr. In the return trip, he covers the speed distance at an average of 30 km/hr. the average speed of the taxi over the entire distance of 680 km is a) 30 b) 40 c) 50 d) 60 

Last Answer : B Average speed =(2xy/( x+y)) km/hr Given x =60 km/hr y = 30km/hr Average speed = (2*60*30/ (60+30)) km/hr =(3600/ 90) =40 km/hr

Rs.1870 is divided into three parts in such a way that half of the first part, one-third of the second part and one-sixth of the third part are equal. The third part is A) Rs.510 B) Rs.680 C) Rs.850 D) Rs.1020 E) None of these

Description : Rs.1870 is divided into three parts in such a way that half of the first part, one-third of the second part and one-sixth of the third part are equal. The third part is A) Rs.510 B) Rs.680 C) Rs.850 D) Rs.1020 E) None of these 

Last Answer : Answer: D F/2 = S/3= T/6 T = 2S = 3F F + S + T = 1870 T( 1/3 +1/2 + 1 ) = 1870 T= 1870 x 6/11 = 170 x 6 = 1020 Or the third part is a multiple of 6 and is large

Two taps can fill a bunker in 40 and 48 mins respectively and a waste tap can empty 6 gallons per minute. All the 3 taps working together can fill the bunker in 30 mind. The capacity of the bunker is. A) 480 B) 480 C) 680 D) 780

Description : Two taps can fill a bunker in 40 and 48 mins respectively and a waste tap can empty 6 gallons per minute. All the 3 taps working together can fill the bunker in 30 mind. The capacity of the bunker is. A) 480 B) 480 C) 680 D) 780 

Last Answer : A Work done by the waste tap in 1 min = 1/30 – (1/40 + 1/48) = 8 – (6+5)/240 = 8-11/240 = -3/240 = -1/80 (negative sign means emptying) Volume of 1/80 part = 60 gallons Volume of whole = (6*80) = 480 gallons

A school has 9 maths teachers and 6 science teachers. In how many ways can a team of 4 maths teachers be formed from them such that the team must contain exactly 1 science teacher? A) 800 B) 720 C) 680 D) 504

Description : A school has 9 maths teachers and 6 science teachers. In how many ways can a team of 4 maths teachers be formed from them such that the team must contain exactly 1 science teacher? A) 800 B) 720 C) 680 D) 504 

Last Answer : Answer: D) The team should have 4 maths teachers. But the team must contain exactly 1 science teacher. Hence, select 3 maths teachers from 9 maths teachers and select 1 science teachers from 6 science teachers.  Number of ways this can be ...  ={9 8 7}/{3 2 1}X6  = 504 / 6 6  = 84 6 =504 

Assertion `:-` In the alveoli, dissociation of `CO_(2)` from carbamino-haemoglobin takes place Reason `:-` In the alveoli `P_(CO_(2))` is low and `P_(

Description : Assertion `:-` In the alveoli, dissociation of `CO_(2)` from carbamino-haemoglobin takes place Reason `:-` In the alveoli `P_(CO_(2))` is low and `P_(

Last Answer : Assertion `:-` In the alveoli, dissociation of `CO_(2)` from carbamino-haemoglobin takes place Reason ` ... . D. If both Assertion & Reason are false.

Assertion `:-` In the tissue high `P_(O_(2))`, low `P_(CO_(2))`, higher `H^(o+)` concentration conditions are favourable for dissociation of oxygen fr

Description : Assertion `:-` In the tissue high `P_(O_(2))`, low `P_(CO_(2))`, higher `H^(o+)` concentration conditions are favourable for dissociation of oxygen fr

Last Answer : Assertion `:-` In the tissue high `P_(O_(2))`, low `P_(CO_(2))`, higher `H^(o+)` ... Reason is False. D. If both Assertion & Reason are false.

In which the following factors can interfere in binding of `O_(2)` with haemoglobin. `A-P_(CO_(2))` `B-H^(+)` ion concentration C-Temperture

Description : In which the following factors can interfere in binding of `O_(2)` with haemoglobin. `A-P_(CO_(2))` `B-H^(+)` ion concentration C-Temperture

Last Answer : In which the following factors can interfere in binding of `O_(2)` with haemoglobin. `A-P_(CO_(2))` `B-H^(+)` ion ... A B. B,C C. A and C D. A,B,C

Write the following steps in sequential order to verify the laws of reflection. (A) Draw a normal at the point where the two lines PQ and `P_(1)Q_(1)`

Description : Write the following steps in sequential order to verify the laws of reflection. (A) Draw a normal at the point where the two lines PQ and `P_(1)Q_(1)`

Last Answer : Write the following steps in sequential order to verify the laws of reflection. (A) Draw a normal at the ... A. DABECF B. ACDBFE C. DECBAF D. DBACEF

Which of the following molecules has both `p_(pi)-p_(pi) and p_(pi)-d_(pi)` bonds

Description : Which of the following molecules has both `p_(pi)-p_(pi) and p_(pi)-d_(pi)` bonds

Last Answer : Which of the following molecules has both `p_(pi)-p_(pi) and p_(pi)-d_(pi)` bonds A. `XeO_(2)F_(2)` B. `SO_(2)Cl_(2)` C. `SO_(3)` D. `XeO_(3)`

What product is formed when acetic acid heated with `P_(2)O_(5)`

Description : What product is formed when acetic acid heated with `P_(2)O_(5)`

Last Answer : What product is formed when acetic acid heated with `P_(2)O_(5)` A. Acetyl chloride B. Acetate ester C. Acetic anhydride D. Acetaldehyde

Nitrogen exists as diatomic molecule and phosphorus as `P_(4)`. Why ?

Description : Nitrogen exists as diatomic molecule and phosphorus as `P_(4)`. Why ?

Last Answer : Nitrogen exists as diatomic molecule and phosphorus as `P_(4)`. Why ? A. `N_(2)` has ... P has preference to adapt structures with small bond angles

`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)`

The nitrogen containing compound produced in the reaction of `HNO_(3)` with `P_(4)O_(10)`

Description : The nitrogen containing compound produced in the reaction of `HNO_(3)` with `P_(4)O_(10)`

Last Answer : The nitrogen containing compound produced in the reaction of `HNO_(3)` with `P_(4)O_(10)` A. ... bond D. react with Na metal producing a brown gas

Among the following, the number of compounds that can react with `PCl_(5)` to give `POCl_(3)` is `O_(2)`,`CO_(2)`,`SO_(2)`,`H_(2)O`,`H_(2)SO_(4)`,`P_(

Description : Among the following, the number of compounds that can react with `PCl_(5)` to give `POCl_(3)` is `O_(2)`,`CO_(2)`,`SO_(2)`,`H_(2)O`,`H_(2)SO_(4)`,`P_(

Last Answer : Among the following, the number of compounds that can react with `PCl_(5)` to give `POCl_(3)` is `O_(2)`,`CO_(2) ... 2)O`,`H_(2)SO_(4)`,`P_(4)O_(10)`.

The reaction of `P_(4)` with X leads selectively to `P_(4)O_(6)` The X is :

Description : The reaction of `P_(4)` with X leads selectively to `P_(4)O_(6)` The X is :

Last Answer : The reaction of `P_(4)` with X leads selectively to `P_(4)O_(6)` The X is : A. Dry `O_(2)` B. ... `O_(2)` D. `O_(2)` in the presence of aqueous NaOH

`N_(2)O_(4), (HPO_(2))_(2), H_(2)CO_(3), SO_(2), SO_(3), P_(4)O_(10) H_(2)SO_(4), N_(2)O_(3), HNO_(3), H_(3)PO_(3)`. (a) Among the above compounds, co

Description : `N_(2)O_(4), (HPO_(2))_(2), H_(2)CO_(3), SO_(2), SO_(3), P_(4)O_(10) H_(2)SO_(4), N_(2)O_(3), HNO_(3), H_(3)PO_(3)`. (a) Among the above compounds, co

Last Answer : `N_(2)O_(4), (HPO_(2))_(2), H_(2)CO_(3), SO_(2), SO_(3), P_(4)O_(10) H_(2)SO_(4), N_(2)O_(3 ... `d pi - p pi` bond are y. Given the answer as `x + y`

The number of O atoms having `sp^(2)` hybridisation in `P_(4)O_(10)` molecule is:

Description : The number of O atoms having `sp^(2)` hybridisation in `P_(4)O_(10)` molecule is:

Last Answer : The number of O atoms having `sp^(2)` hybridisation in `P_(4)O_(10)` molecule is:

Number of gaseous oxides among the following at room temperature is: (a) `N_(2)O` (b) `NO` (c) `N_(2)O_(3)` (d) `NO_(2)` (e) `N_(2)O_(5)` (f) `P_(4)O_

Description : Number of gaseous oxides among the following at room temperature is: (a) `N_(2)O` (b) `NO` (c) `N_(2)O_(3)` (d) `NO_(2)` (e) `N_(2)O_(5)` (f) `P_(4)O_

Last Answer : Number of gaseous oxides among the following at room temperature is: (a) `N_(2)O` (b) `NO` (c) `N_(2)O_(3)` ( ... (4)O_(10)` (h) `SO_(2)` (i) `SO_(3)`

Number of acidic oxides among the following is: (a) `N_(2)O` (b) `NO` (c) `N_(2)O_(3)` (d) `N_(2)O_(4)` (e) `N_(2)O_(5)` (f) `P_(4)O_(6)` (g) `P_(4)O_

Description : Number of acidic oxides among the following is: (a) `N_(2)O` (b) `NO` (c) `N_(2)O_(3)` (d) `N_(2)O_(4)` (e) `N_(2)O_(5)` (f) `P_(4)O_(6)` (g) `P_(4)O_

Last Answer : Number of acidic oxides among the following is: (a) `N_(2)O` (b) `NO` (c) `N_(2)O_(3)` (d) `N_(2)O_(4)` (e ... (h) `SO_(3)` (i) `B_(2)O_(3)` (j) `CO`

When `P_(4)O_(10)` is dissolved in water, the acid formed finally is :

Description : When `P_(4)O_(10)` is dissolved in water, the acid formed finally is :

Last Answer : When `P_(4)O_(10)` is dissolved in water, the acid formed finally is : A. `H_(3)PO_(2)` B. `H_(3)PO_(4)` C. `H_(3)PO_(3)` D. `H_(4)P_(2)O_(7)`

`HNO_(3)+P_(4)O_(10) to HPO_(3) + X` in the above reaction the product `X` is :

Description : `HNO_(3)+P_(4)O_(10) to HPO_(3) + X` in the above reaction the product `X` is :

Last Answer : `HNO_(3)+P_(4)O_(10) to HPO_(3) + X` in the above reaction the product `X` is : A. `NO_(2)` B. `N_(2)O_(3)` C. `N_(2)O_(4)` D. `N_(2)O_(5)`

What product is formed when acetic acid heated with `P_(2)O_(5)`

Description : What product is formed when acetic acid heated with `P_(2)O_(5)`

Last Answer : What product is formed when acetic acid heated with `P_(2)O_(5)` A. Acetyl chloride B. Acetate ester C. Acetic anhydride D. Acetaldehyde

Nitrogen exists as diatomic molecule and phosphorus as `P_(4)`. Why ?

Description : Nitrogen exists as diatomic molecule and phosphorus as `P_(4)`. Why ?

Last Answer : Nitrogen exists as diatomic molecule and phosphorus as `P_(4)`. Why ? A. `N_(2)` has ... P has preference to adapt structures with small bond angles

`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)`

The nitrogen containing compound produced in the reaction of `HNO_(3)` with `P_(4)O_(10)`

Description : The nitrogen containing compound produced in the reaction of `HNO_(3)` with `P_(4)O_(10)`

Last Answer : The nitrogen containing compound produced in the reaction of `HNO_(3)` with `P_(4)O_(10)` A. ... bond D. react with Na metal producing a brown gas

Among the following, the number of compounds that can react with `PCl_(5)` to give `POCl_(3)` is `O_(2)`,`CO_(2)`,`SO_(2)`,`H_(2)O`,`H_(2)SO_(4)`,`P_(

Description : Among the following, the number of compounds that can react with `PCl_(5)` to give `POCl_(3)` is `O_(2)`,`CO_(2)`,`SO_(2)`,`H_(2)O`,`H_(2)SO_(4)`,`P_(

Last Answer : Among the following, the number of compounds that can react with `PCl_(5)` to give `POCl_(3)` is `O_(2)`,`CO_(2) ... 2)O`,`H_(2)SO_(4)`,`P_(4)O_(10)`.

The reaction of `P_(4)` with X leads selectively to `P_(4)O_(6)` The X is :

Description : The reaction of `P_(4)` with X leads selectively to `P_(4)O_(6)` The X is :

Last Answer : The reaction of `P_(4)` with X leads selectively to `P_(4)O_(6)` The X is : A. Dry `O_(2)` B. ... `O_(2)` D. `O_(2)` in the presence of aqueous NaOH

`N_(2)O_(4), (HPO_(2))_(2), H_(2)CO_(3), SO_(2), SO_(3), P_(4)O_(10) H_(2)SO_(4), N_(2)O_(3), HNO_(3), H_(3)PO_(3)`. (a) Among the above compounds, co

Description : `N_(2)O_(4), (HPO_(2))_(2), H_(2)CO_(3), SO_(2), SO_(3), P_(4)O_(10) H_(2)SO_(4), N_(2)O_(3), HNO_(3), H_(3)PO_(3)`. (a) Among the above compounds, co

Last Answer : `N_(2)O_(4), (HPO_(2))_(2), H_(2)CO_(3), SO_(2), SO_(3), P_(4)O_(10) H_(2)SO_(4), N_(2)O_(3 ... `d pi - p pi` bond are y. Given the answer as `x + y`

The number of O atoms having `sp^(2)` hybridisation in `P_(4)O_(10)` molecule is:

Description : The number of O atoms having `sp^(2)` hybridisation in `P_(4)O_(10)` molecule is:

Last Answer : The number of O atoms having `sp^(2)` hybridisation in `P_(4)O_(10)` molecule is:

Number of gaseous oxides among the following at room temperature is: (a) `N_(2)O` (b) `NO` (c) `N_(2)O_(3)` (d) `NO_(2)` (e) `N_(2)O_(5)` (f) `P_(4)O_

Description : Number of gaseous oxides among the following at room temperature is: (a) `N_(2)O` (b) `NO` (c) `N_(2)O_(3)` (d) `NO_(2)` (e) `N_(2)O_(5)` (f) `P_(4)O_

Last Answer : Number of gaseous oxides among the following at room temperature is: (a) `N_(2)O` (b) `NO` (c) `N_(2)O_(3)` ( ... (4)O_(10)` (h) `SO_(2)` (i) `SO_(3)`

Number of acidic oxides among the following is: (a) `N_(2)O` (b) `NO` (c) `N_(2)O_(3)` (d) `N_(2)O_(4)` (e) `N_(2)O_(5)` (f) `P_(4)O_(6)` (g) `P_(4)O_

Description : Number of acidic oxides among the following is: (a) `N_(2)O` (b) `NO` (c) `N_(2)O_(3)` (d) `N_(2)O_(4)` (e) `N_(2)O_(5)` (f) `P_(4)O_(6)` (g) `P_(4)O_

Last Answer : Number of acidic oxides among the following is: (a) `N_(2)O` (b) `NO` (c) `N_(2)O_(3)` (d) `N_(2)O_(4)` (e ... (h) `SO_(3)` (i) `B_(2)O_(3)` (j) `CO`

When `P_(4)O_(10)` is dissolved in water, the acid formed finally is :

Description : When `P_(4)O_(10)` is dissolved in water, the acid formed finally is :

Last Answer : When `P_(4)O_(10)` is dissolved in water, the acid formed finally is : A. `H_(3)PO_(2)` B. `H_(3)PO_(4)` C. `H_(3)PO_(3)` D. `H_(4)P_(2)O_(7)`

`HNO_(3)+P_(4)O_(10) to HPO_(3) + X` in the above reaction the product `X` is :

Description : `HNO_(3)+P_(4)O_(10) to HPO_(3) + X` in the above reaction the product `X` is :

Last Answer : `HNO_(3)+P_(4)O_(10) to HPO_(3) + X` in the above reaction the product `X` is : A. `NO_(2)` B. `N_(2)O_(3)` C. `N_(2)O_(4)` D. `N_(2)O_(5)`

The unbalanced equation for the reaction of `P_(4)S_(3)` with nitrate in aqueous acidic medium is given below : `P_(4)S_(3)+NO_(3)^(-)rarrH_(3)PO_(4)+

Description : The unbalanced equation for the reaction of `P_(4)S_(3)` with nitrate in aqueous acidic medium is given below : `P_(4)S_(3)+NO_(3)^(-)rarrH_(3)PO_(4)+

Last Answer : The unbalanced equation for the reaction of `P_(4)S_(3)` with nitrate in aqueous acidic medium is given below : `P_(4) ... A. 18 B. `8//3` C. 8 D. 28