How many ml in 10x10x10cm volume?

How many ml in 10x10x10cm volume?
by

1 Answer

Answer :

There are 1000 millilitres.
by
selected by

Related questions

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

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

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

A 500 mL glass flask is filled at 298 K and 1 atm, pressure with three diatomic gases, X, Y and Z . The initial volume ratio of the gases before mixin

Description : A 500 mL glass flask is filled at 298 K and 1 atm, pressure with three diatomic gases, X, Y and Z . The initial volume ratio of the gases before mixin

Last Answer : A 500 mL glass flask is filled at 298 K and 1 atm, pressure with three diatomic gases, X, Y and Z . The ... . `H_2 , F_2 , O_2` D. `O_2,H_2 , F_2`

What volume of water is requried to make `0.20 N` solution from `1600 mL` of `0.2050 N` solution?

Description : What volume of water is requried to make `0.20 N` solution from `1600 mL` of `0.2050 N` solution?

Last Answer : What volume of water is requried to make `0.20 N` solution from `1600 mL` of `0.2050 N` solution? A. 40 mL B ... to `H_(3)AsO_(4)` C. 100 mL D. 20 mL

25 ml of the given HCl solution requires 30 mL of 0.1M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of

Description : 25 ml of the given HCl solution requires 30 mL of 0.1M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of

Last Answer : 25 ml of the given HCl solution requires 30 mL of 0.1M sodium carbonate solution. What is the volume of this HCl ... 5 mL B. 75 mL C. 50 mL D. 25 mL

Potassium acid oxalate `K_(2)C_(2)O_(4).3H_(2)C_(2)O_(4).4H_(2)O` can be oxidized by `MnO_(4)^(-)` in acid medium. Calculate the volume of (in mL) 1 M

Description : Potassium acid oxalate `K_(2)C_(2)O_(4).3H_(2)C_(2)O_(4).4H_(2)O` can be oxidized by `MnO_(4)^(-)` in acid medium. Calculate the volume of (in mL) 1 M

Last Answer : Potassium acid oxalate `K_(2)C_(2)O_(4).3H_(2)C_(2)O_(4).4H_(2)O` can be oxidized by ... reacting in acid solution with 5.08 gram of the acid oxalate.

The volume of 3 M `Ba(OH)^(2)` solution required to neutralize completely 120 mL of 1.5M `H_(3)PO_(4)` solution is :

Description : The volume of 3 M `Ba(OH)^(2)` solution required to neutralize completely 120 mL of 1.5M `H_(3)PO_(4)` solution is :

Last Answer : The volume of 3 M `Ba(OH)^(2)` solution required to neutralize completely 120 mL of 1.5M `H_(3)PO_(4)` solution is :

The volume (in `mL`) of `0.1M Ag NO_(3)` required for complete precipitation of chloride ions present in `30 mL` of `0.01M` solution of `[Cr(H_(2)O)_(

Description : The volume (in `mL`) of `0.1M Ag NO_(3)` required for complete precipitation of chloride ions present in `30 mL` of `0.01M` solution of `[Cr(H_(2)O)_(

Last Answer : The volume (in `mL`) of `0.1M Ag NO_(3)` required for complete precipitation of chloride ions present ... Cl]Cl_(2)`, as silver chloride is close to:

Co-ordination number of Cr in `CrCl_(3^.)5H_(2)O` is six. The volume of 0.1 N `AgNO_(3)` needed to ppt. the chorine in outer sphere in 200 ml of 0.01

Description : Co-ordination number of Cr in `CrCl_(3^.)5H_(2)O` is six. The volume of 0.1 N `AgNO_(3)` needed to ppt. the chorine in outer sphere in 200 ml of 0.01

Last Answer : Co-ordination number of Cr in `CrCl_(3^.)5H_(2)O` is six. The volume of 0.1 N `AgNO_(3)` ... of 0.01 M solution of he possible complex es is/are:

At 25°C and 760 mm of Hg pressure a gas occupies 600 mL volume.

Description : At 25°C and 760 mm of Hg pressure a gas occupies 600 mL volume.

Last Answer : At 25°C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at ... temperature is 10°C and volume of the gas 640 mL ?

If A cup as sugar has a volume of 237 ml what is the maa o the cup if the cup of sugar if the density is 1.59 g ml?

Description : If A cup as sugar has a volume of 237 ml what is the maa o the cup if the cup of sugar if the density is 1.59 g ml?

Last Answer : Need answer

What element has a mass of 170.1 and a volume of 16.2 ml what material is it?

Description : What element has a mass of 170.1 and a volume of 16.2 ml what material is it?

Last Answer : Need answer

If you were asked to add 50 mL of a stock solution to a reaction vessel which piece of glassware would you use to obtain this volume of solution?

Description : If you were asked to add 50 mL of a stock solution to a reaction vessel which piece of glassware would you use to obtain this volume of solution?

Last Answer : You can use a graduated cylinder or a pipette of 50 mL.

What volume (in mL) of 0.25M NaSO4 solution is needed to precipitate all the barium as BaSO4(s) from 12.5mL of 0.15M Ba(NO3)2 solution?

Description : What volume (in mL) of 0.25M NaSO4 solution is needed to precipitate all the barium as BaSO4(s) from 12.5mL of 0.15M Ba(NO3)2 solution?

Last Answer : The chemical equation is:Ba(NO3)2 + Na2SO4 = BaSO4 + 2 NaNO3The volume (in mL) of 0,25M Na2SO4 solution needed to precipitate all the barium as BaSO4(s) from 12,5mL of 0,15M Ba(NO3)2 solution is 7,5 mL.

What volume in mL of a 0.10 M NaOH solution would be required to neutralize 0.0091 moles of CH3COOH in a given vinegar sample?

Description : What volume in mL of a 0.10 M NaOH solution would be required to neutralize 0.0091 moles of CH3COOH in a given vinegar sample?

Last Answer : The volume of NaOH is 99 mL.

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

Minimum excretory urinary volume for waste products elimination during 24 hrs is (A) 200–300 ml (B) 200–400 ml (C) 500–600 ml (D) 800 ml

Description : Minimum excretory urinary volume for waste products elimination during 24 hrs is (A) 200–300 ml (B) 200–400 ml (C) 500–600 ml (D) 800 ml

Last Answer : Answer : C

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

ADH test is based on the measurement of (A) Specific gravity of urine (B) Concentration of urea in urine (C) Concentration of urea in blood (D) Volume of urine in ml/minute

Description : ADH test is based on the measurement of (A) Specific gravity of urine (B) Concentration of urea in urine (C) Concentration of urea in blood (D) Volume of urine in ml/minute

Last Answer : A

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

The number of water molecules present in a drop of water (volume 0.0018 ml) at room temperature is?

Description : The number of water molecules present in a drop of water (volume 0.0018 ml) at room temperature is?

Last Answer : 6.023 x 1019

The cerebrospinal fluid: a. has a normal volume of 150 ml b. has a normal opening pressure of 7 - 18 cm H2O c. flows from the ventricles to the subarachnoid space via the foramen of Monrod. does not contain neutrophils in normal individuals

Description : The cerebrospinal fluid: a. has a normal volume of 150 ml b. has a normal opening pressure of 7 - 18 cm H2O c. flows from the ventricles to the subarachnoid space via the foramen of Monro d. does not contain neutrophils in normal individuals

Last Answer : has a normal volume of 150 m

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

When 1500 mL air is in the lungs, it is called (a) residual volume (b) inspiratory reserve volume (c) vital capacity (d) tidal volume.

Description : When 1500 mL air is in the lungs, it is called (a) residual volume (b) inspiratory reserve volume (c) vital capacity (d) tidal volume.

Last Answer : (a) residual volume

Listed below are four respiratory capacities (i–iv) and four jumbled respiratory volumes of a normal human adult. Respiratory Respiratory capacities volumes (i) Residual volume 2500 mL (ii) Vital capacity 3500 mL (iii)Inspiratory reserve volume 1200 mL (iv) Inspiratory capacity 4500 mL Which one of the following is the correct matching of two capacities and volumes? (a) (ii) 2500 mL, (iii) 4500 mL (b) (iii) 1200 mL, (iv) 2500 mL (c) (iv) 3500 mL, (i) 1200 mL (d) (i) 4500 mL, (ii) 3500 mL

Description : Listed below are four respiratory capacities (i-iv) and four jumbled respiratory volumes of a normal human adult. Respiratory Respiratory capacities volumes (i) Residual volume 2500 mL (ii) Vital capacity 3500 mL (iii)Inspiratory ... c) (iv) 3500 mL, (i) 1200 mL (d) (i) 4500 mL, (ii) 3500 mL

Last Answer : (c) (iv) 3500 mL, (i) 1200 mL

Match the items given in column I with those in column II and select the correct option given below. Column I Column II (A) Tidal volume (i) 2500 – 3000 mL (B) Inspiratory reserve (ii) 1100 – 1200 mL volume (C) Expiratory reserve (iii) 500 – 550 mL volume (D) Residual volume (iv) 1000 – 1100 mL (A) (B) (C) (D) (a) (iii) (ii) (i) (iv) (b) (iii) (i) (iv) (ii) (c) (i) (iv) (ii) (iii) (d) (iv) (iii) (ii) (i)

Description : Match the items given in column I with those in column II and select the correct option given below. Column I Column II (A) Tidal volume (i) 2500 - 3000 mL (B) Inspiratory reserve (ii) 1100 - 1200 mL volume (C) Expiratory reserve (iii ... i) (iv) (ii) (c) (i) (iv) (ii) (iii) (d) (iv) (iii) (ii) (i)

Last Answer : (b) (iii) (i) (iv) (ii)

Tidal volume and expiratory reserve volume of an athlete is 500 mL and 1000 mL respectively. What will be his expiratory capacity if the residual volume is 1200 mL? (a) 2700 mL (b) 1500 mL (c) 1700 mL (d) 2200 mL

Description : Tidal volume and expiratory reserve volume of an athlete is 500 mL and 1000 mL respectively. What will be his expiratory capacity if the residual volume is 1200 mL? (a) 2700 mL (b) 1500 mL (c) 1700 mL (d) 2200 mL

Last Answer : b) 1500 mL

When the nurse notes that the post cardiac surgery patient demonstrates low urine output (< 25 ml/hr) with high specific gravity (> 1.025), the nurse suspects: a) Inadequate fluid volume Urine output of less than 25 ml/hr may indicate a decrease in cardiac output. A high specific gravity indicates increased concentration of solutes in the urine which occurs with inadequate fluid volume. b) Normal glomerular filtration Indices of normal glomerular filtration are output of 25 ml or greater per hour and specific gravity between 1.010 and 1.025. c) Overhydration Overhydration is manifested by high urine output with low specific gravity. d) Anuria The anuric patient does not produce urine.

Description : When the nurse notes that the post cardiac surgery patient demonstrates low urine output (< 25 ml/hr) with high specific gravity (> 1.025), the nurse suspects: a) Inadequate fluid volume Urine ... by high urine output with low specific gravity. d) Anuria The anuric patient does not produce urine.

Last Answer : a) Inadequate fluid volume Urine output of less than 25 ml/hr may indicate a decrease in cardiac output. A high specific gravity indicates increased concentration of solutes in the urine which occurs with inadequate fluid volume.

Which respiratory volume is the maximum volume of air that can be inhaled after maximal expiration? a) Inspiratory reserve volume Inspiratory reserve volume is normally 3000 mL. b) Tidal volume Tidal volume is the volume of air inhaled and exhaled with each breath. c) Expiratory reserve volume Expiratory reserve volume is the maximum volume of air that can be exhaled forcibly after a normal exhalation. d) Residual volume Residual volume is the volume of air remaining in the lungs after a maximum exhalation.

Description : Which respiratory volume is the maximum volume of air that can be inhaled after maximal expiration? a) Inspiratory reserve volume Inspiratory reserve volume is normally 3000 mL. b) Tidal volume Tidal ... volume Residual volume is the volume of air remaining in the lungs after a maximum exhalation.

Last Answer : a) Inspiratory reserve volume Inspiratory reserve volume is normally 3000 mL.

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