how do you convert 0.25mL of 0.10M HCI → mols HCI

how do you convert 0.25mL of 0.10M HCI → mols HCI
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Which of the following samples of reducing agents is /are chemically equivalent to 25mL of 0.2 N `KMnO_(4)` to be reduced to `Mn^(2+)` and water?

Description : Which of the following samples of reducing agents is /are chemically equivalent to 25mL of 0.2 N `KMnO_(4)` to be reduced to `Mn^(2+)` and water?

Last Answer : Which of the following samples of reducing agents is /are chemically equivalent to 25mL of 0.2 N `KMnO_(4)` ... SnCl_(2)` to be oxidized to `Sn^(4+)`

`25ml` of a `0.1(M)` solution of a stable cation of transition metal `z` reacts exactly with `25 ml` of `0.04(M)` acidified `KMnO_(4)` solution. Which

Description : `25ml` of a `0.1(M)` solution of a stable cation of transition metal `z` reacts exactly with `25 ml` of `0.04(M)` acidified `KMnO_(4)` solution. Which

Last Answer : `25ml` of a `0.1(M)` solution of a stable cation of transition metal `z` reacts exactly with `25 ml` of `0.04 ... 3+)rarrZ^(4+)` D. `Z^(2+)rarrZ^(4+)`

Which of the following solutions will exactly oxidize `25mL` of an acid solution of `0.1 M Fe` (`II`) oxalate?

Description : Which of the following solutions will exactly oxidize `25mL` of an acid solution of `0.1 M Fe` (`II`) oxalate?

Last Answer : Which of the following solutions will exactly oxidize `25mL` of an acid solution of `0.1 M Fe` (`II`) oxalate? A. ... 4)` D. 15 mL of 0.1 M `KMnO_(4)`

How many mols in 74g of KCl?

Description : How many mols in 74g of KCl?

Last Answer : 1 mole.

How many mols in 100g of KMnO4?

Description : How many mols in 100g of KMnO4?

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One mol. of Urea is synthesized at the expense of the _______ mols. of ATP. (A) 2 (B) 3 (C) 4 (D) 5

Description : One mol. of Urea is synthesized at the expense of the _______ mols. of ATP. (A) 2 (B) 3 (C) 4 (D) 5

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Out of 24 mols of ATP formed in TCA cycle, 2 molecules of ATP can be formed at “substrate level” by which of the following reaction ? (A) Citric acid→ Isocitric acid (B) Isocitrate→ Oxaloacetate (C) Succinic acid→ Fumarate (D) Succinylcat→ Succinic acid

Description : Out of 24 mols of ATP formed in TCA cycle, 2 molecules of ATP can be formed at “substrate level” by which of the following reaction ? (A) Citric acid→ Isocitric acid (B) Isocitrate→ Oxaloacetate (C) Succinic acid→ Fumarate (D) Succinylcat→ Succinic acid

Last Answer : D

If 25mL of a `H_(2)SO_(4)` solution reacts completely with `1.06g` of pure `Na_(2)CO_(3)` , what is the normality of this acid sotution :

Description : If 25mL of a `H_(2)SO_(4)` solution reacts completely with `1.06g` of pure `Na_(2)CO_(3)` , what is the normality of this acid sotution :

Last Answer : If 25mL of a `H_(2)SO_(4)` solution reacts completely with `1.06g` of pure `Na_(2)CO_(3)` , what is the normality of ... N B. 0.5 N C. 1.8 N D. 0.8 N

How many ounces in 25ml?

Description : How many ounces in 25ml?

Last Answer : What is the answer ?

If we convert ∃u ∀v ∀x ∃y (P(f(u),v, x, y) → Q(u,v,y)) to ∀v ∀x (P(f(a),v, x, g(v,x)) → Q(a,v,g(v,x))) This process is known as (A) Simplification (B) Unification (C) Skolemization (D) Resolution

Description : If we convert ∃u ∀v ∀x ∃y (P(f(u),v, x, y) → Q(u,v,y)) to ∀v ∀x (P(f(a),v, x, g(v,x)) → Q(a,v,g(v,x))) This process is known as (A) Simplification (B) Unification (C) Skolemization (D) Resolution

Last Answer : (C) Skolemization

A uniform rope of mass M=0.1kg and length L=10m hangs from the celling. `[g=10m//s^(2)]` :-

Description : A uniform rope of mass M=0.1kg and length L=10m hangs from the celling. `[g=10m//s^(2)]` :-

Last Answer : A uniform rope of mass M=0.1kg and length L=10m hangs from the celling. `[g=10m//s^(2) ... in the rope remain constant along the length of the rope

4. A three-phase, four-wire system having \( 250 V \) line-to-neutral has the following loads connected between the respective lines and the neutral. Calculate the current in the neutral wire?\[Z_{R}=20 \angle 0^{\circ} \Omega ; Z_{Y}=20 \angle 35^{\circ} \Omega, Z_{B}=20 \angle-55^{\circ} \Omega\](10M)

Description : 4. A three-phase, four-wire system having \( 250 V \) line-to-neutral has the following loads connected between the respective lines and the neutral. Calculate the current in the neutral wire?\[Z_{R}=20 \angle 0^{\circ} ... Z_{Y}=20 \angle 35^{\circ} \Omega, Z_{B}=20 \angle-55^{\circ} \Omega\](10M)

Last Answer : 4. A three-phase, four-wire system having \( 250 V \) line-to-neutral has the following loads connected ... =20 \angle-55^{\circ} \Omega \] (10M)

4. A three-phase, four-wire system having \( 250 V \) line-to-neutral has the following loads connected between the respective lines and the neutral. Calculate the current in the neutral wire?\[Z_{R}=20 \angle 0^{\circ} \Omega ; Z_{Y}=20 \angle 35^{\circ} \Omega, Z_{B}=20 \angle-55^{\circ} \Omega\](10M)

Description : 4. A three-phase, four-wire system having \( 250 V \) line-to-neutral has the following loads connected between the respective lines and the neutral. Calculate the current in the neutral wire?\[Z_{R}=20 \angle 0^{\circ} ... Z_{Y}=20 \angle 35^{\circ} \Omega, Z_{B}=20 \angle-55^{\circ} \Omega\](10M)

Last Answer : 4. A three-phase, four-wire system having \( 250 V \) line-to-neutral has the following loads connected ... =20 \angle-55^{\circ} \Omega \] (10M)

4. A three-phase, four-wire system having \( 250 V \) line-to-neutral has the following loads connected between the respective lines and the neutral. Calculate the current in the neutral wire?\[Z_{R}=20 \angle 0^{\circ} \Omega ; Z_{Y}=20 \angle 35^{\circ} \Omega, Z_{B}=20 \angle-55^{\circ} \Omega\](10M)

Description : 4. A three-phase, four-wire system having \( 250 V \) line-to-neutral has the following loads connected between the respective lines and the neutral. Calculate the current in the neutral wire?\[Z_{R}=20 \angle 0^{\circ} ... Z_{Y}=20 \angle 35^{\circ} \Omega, Z_{B}=20 \angle-55^{\circ} \Omega\](10M)

Last Answer : 4. A three-phase, four-wire system having \( 250 V \) line-to-neutral has the following loads connected ... =20 \angle-55^{\circ} \Omega \] (10M)

Which of the following is closest to a yard in length? A. 0.01m B. 0.1m C. 1m (Answer) D. 10m E. 100m

Description : Which of the following is closest to a yard in length? A. 0.01m B. 0.1m C. 1m (Answer) D. 10m E. 100m

Last Answer : C. 1m (Answer)

The electric field intensity of a field with velocity 10m/s and flux density of 2.8 units is a) 0.28 b) 28 c) 280 d) 10/2.8

Description : The electric field intensity of a field with velocity 10m/s and flux density of 2.8 units is a) 0.28 b) 28 c) 280 d) 10/2.8

Last Answer : b) 28

A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water vapor at 100 kPa. Calculate the internal energy of the system using the steam table.  A. 5 x10^5 kJ  B. 8x10^5 kJ  C. 1 x10^6 kJ  D. 2 x10^6 kJ Formula: fromthe steamtable vƒ = 0.001043 m^3 / kg vg = 1.6940 m^3 / kg u ƒ= 417.3 kJ/kg ug= 2506kJ/kg formula: Mvap = V vap/vg M liq = Vliq/ vƒ u =uƒM liq + ug M vap

Description : A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water vapor at 100 kPa. Calculate the internal energy of the system using the steam table.  A. 5 x10^5 kJ  B. 8x10^5 kJ  C. 1 ... 3 kJ/kg ug= 2506kJ/kg formula: Mvap = V vap/vg M liq = Vliq/ vƒ u =uƒM liq + ug M vap

Last Answer : 2 x10^6 kJ

A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml water and titrated with 0.5 M HCI when of `((1)/(5))^(th

Description : A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml water and titrated with 0.5 M HCI when of `((1)/(5))^(th

Last Answer : A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml ... D. Weight percentage of base in given sample is 80%.

When an primary aromatic amine is treated with `NaNO_(2)+HCI` at `0^(@)-5^(@)C`, a diazonium salt is formed and the reaction is called diazo reaction.

Description : When an primary aromatic amine is treated with `NaNO_(2)+HCI` at `0^(@)-5^(@)C`, a diazonium salt is formed and the reaction is called diazo reaction.

Last Answer : When an primary aromatic amine is treated with `NaNO_(2)+HCI` at `0^(@)-5^(@)C`, a diazonium salt is ... is formed which can be given as: A. B. C. D.

When an primary aromatic amine is treated with `NaNO_(2)+HCI` at `0^(@)-5^(@)C`, a diazonium salt is formed and the reaction is called diazo reaction.

Description : When an primary aromatic amine is treated with `NaNO_(2)+HCI` at `0^(@)-5^(@)C`, a diazonium salt is formed and the reaction is called diazo reaction.

Last Answer : When an primary aromatic amine is treated with `NaNO_(2)+HCI` at `0^(@)-5^(@)C`, a diazonium salt ... reaction. The final product is A. B. C. D.

When an primary aromatic amine is treated with `NaNO_(2)+HCI` at `0^(@)-5^(@)C`, a diazonium salt is formed and the reaction is called diazo reaction.

Description : When an primary aromatic amine is treated with `NaNO_(2)+HCI` at `0^(@)-5^(@)C`, a diazonium salt is formed and the reaction is called diazo reaction.

Last Answer : When an primary aromatic amine is treated with `NaNO_(2)+HCI` at `0^(@)-5^(@)C`, a diazonium salt is formed and ... gt R gt S` D. `S gt R gt Q gt P`

The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Description : The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Last Answer : Inner radius of circular well, r = 3.5/2m = 1.75m Depth of circular well, say h = 10m (i) Inner curved surface area = 2πrh = (2 (22/7 ) 1.75 10) = 110 Therefore, the inner curved surface ... area = Rs (110 40) = Rs.4400 Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

The length of the longest pole that can be put in a room of dimensions (10m x 10m x 5m) is -Maths 9th

Description : The length of the longest pole that can be put in a room of dimensions (10m x 10m x 5m) is -Maths 9th

Last Answer : longest pole can be faced along the diagonal and its length =102+102+52​=225​=15m

The length of the longest pole that can be put in a room of dimensions (10m x 10m x 5m) is -Maths 9th

Description : The length of the longest pole that can be put in a room of dimensions (10m x 10m x 5m) is -Maths 9th

Last Answer : Solution of this question

ABCD is a parallelogram AE pependicular to DC CF perpendixular to AD AB =16 m ,AE =8m ,CF =10m ,fimd AD -Maths 9th

Description : ABCD is a parallelogram AE pependicular to DC CF perpendixular to AD AB =16 m ,AE =8m ,CF =10m ,fimd AD -Maths 9th

Last Answer : This answer was deleted by our moderators...

Is 10m greater than 120cm?

Description : Is 10m greater than 120cm?

Last Answer : yes10.0m - 1.2m = 8.8m

What is the area of a cylinder of 1m x 10m?

Description : What is the area of a cylinder of 1m x 10m?

Last Answer : First, what do you mean by "area of a cylinder"?Area is a property of two dimensional shapes whereas a cylinder is a three dimensional shape.I guess that you mean "surface area of a cylinder"Next, you ... , 10m = height1m = diameter, 10m = height1m = height, 10m = radius1m = height, 10m = diameter.

Calculate the electric field when the conductivity is 20 units, electron density is 2.4 units and the velocity is 10m/s. Assume the conduction and convection current densities are same. a) 2.4 b) 4.8 c) 3.6 d) 1.2

Description : Calculate the electric field when the conductivity is 20 units, electron density is 2.4 units and the velocity is 10m/s. Assume the conduction and convection current densities are same. a) 2.4 b) 4.8 c) 3.6 d) 1.2

Last Answer : d) 1.2

The resistance of a material with conductivity 2millimho/m 2 , length 10m and area 50m is a) 500 b) 200 c) 100 d) 1000

Description : The resistance of a material with conductivity 2millimho/m 2 , length 10m and area 50m is a) 500 b) 200 c) 100 d) 1000

Last Answer : c) 100

The resistivity of a material with resistance 200 ohm, length 10m and area twice that of the length is a) 200 b) 300 c) 400 d) 500

Description : The resistivity of a material with resistance 200 ohm, length 10m and area twice that of the length is a) 200 b) 300 c) 400 d) 500

Last Answer : c) 400

A bayabas falls from a branch 5m above the ground with what speed in meter per second does it strike the ground assume g = 10m/s².  a. 11m/s  b. 12m/s  c. 13m/s  d. 10m/s ∫KE = mV2/2gc

Description : A bayabas falls from a branch 5m above the ground with what speed in meter per second does it strike the ground assume g = 10m/s².  a. 11m/s  b. 12m/s  c. 13m/s  d. 10m/s ∫KE = mV2/2gc

Last Answer : 10m/s

A,B are start walking around a Circular park of radius 70m. They start at the same point, and A goes clockwise at 10m/s, while b goes anti-clockwise at 20m/s.How many times will they cross each other at the starting point if they walk for 30 minutes? a) 20 b) 40 c) 60 d) 80 e) None of the above

Description : A,B are start walking around a Circular park of radius 70m. They start at the same point, and A goes clockwise at 10m/s, while b goes anti-clockwise at 20m/s.How many times will they cross each other at the starting point if they walk for 30 minutes? a) 20 b) 40 c) 60 d) 80 e) None of the above

Last Answer : Radius = 70m, So circumference = 2 (22/7) 70 =440m Relative speed = 10 +20 = 30m/s They will meet everytime at the starting point when the distance covered by both put together is a multiple of 440 ... =1800/44 =40.9 So they will meet 40 times (we need to ignore the decimal part). Answer : b

A,B are start walking around a Circular park of radius 70m. They start at the same point, and A goes clockwise at 10m/s, while b goes anti-clockwise at 20m/s.How many times will they cross each other at the starting point if they walk for 30 minutes? a) 20 b) 40 c) 60 d) 80 e) None of the above

Description : A,B are start walking around a Circular park of radius 70m. They start at the same point, and A goes clockwise at 10m/s, while b goes anti-clockwise at 20m/s. How many times will they cross each other at the starting point if they walk for 30 minutes? a) 20 b) 40 c) 60 d) 80 e) None of the above

Last Answer : Radius = 70m, So circumference = 2 (22/7) 70 =440m Relative speed = 10 +20 = 30m/s They will meet everytime at the starting point when the distance covered by both put together is a multiple of 440 ... =1800/44 =40.9 So they will meet 40 times (we need to ignore the decimal part). Answer : b

In racing over a given distance, A can beat B by 20m, B can beat C by 10m, A can beat C by 28m. The racing distance is? a) 58 m b) 116 m c) 100 m d) 120 m e) cannot be determined

Description : In racing over a given distance, A can beat B by 20m, B can beat C by 10m, A can beat C by 28m. The racing distance is? a) 58 m b) 116 m c) 100 m d) 120 m e) cannot be determined

Last Answer : Let the uniform speeds of A,B,C be vA,vB,vC, respectively. Let t1 be the time taken for the race between A and B and t2 be the time taken for race between B and C Since A covers the distance in time t1, race between A and C ... ) = VB/ VC = d/(d-10) Thus (d-20)(d-10)=d(d-28).Hence, d=100. Answer: c)

Basic values of span/effective depth ratios to be used for beams and slabs with spans [ A ] Less than 10m [ B ] Less than 20 m [ C ] 10 m- 20 m [ D ] More than 20 m

Description : Basic values of span/effective depth ratios to be used for beams and slabs with spans [ A ] Less than 10m [ B ] Less than 20 m [ C ] 10 m- 20 m [ D ] More than 20 m

Last Answer : [ A ] Less than 10m

The vertical spacing between ties must be no more than? a. 8m (26 feet) b. 10m (33 feet) c. 12m (39 feet) d. 6m (20 feet)

Description : The vertical spacing between ties must be no more than? a. 8m (26 feet) b. 10m (33 feet) c. 12m (39 feet) d. 6m (20 feet)

Last Answer : a. 8m (26 feet)

The standard length of the shaft is (a) 5 m (b) 8 m (c) 9 m (d) 10m

Description : The standard length of the shaft is (a) 5 m (b) 8 m (c) 9 m (d) 10m

Last Answer : (a) 5 m

eripheral velocity of impact cutter is about: A.50m/sec B.20m/sec C.10m/sec D.5m/sec

Description : eripheral velocity of impact cutter is about: A.50m/sec B.20m/sec C.10m/sec D.5m/sec

Last Answer : A.50m/sec

While doing final location survey, the centre line is fully marked with pegs at --- metre distance (a) 10m (b) 20m* (c) 30m (d) 50m

Description : While doing final location survey, the centre line is fully marked with pegs at --- metre distance (a) 10m (b) 20m* (c) 30m (d) 50m

Last Answer : (b) 20m*

A bridge is generally specified as “Br. No. 180 (5x9.10m G) at km 345/5-6”. a) Effective Span b) Width of bridge c) Clear Span* d) Overall length of bridge

Description : A bridge is generally specified as “Br. No. 180 (5x9.10m G) at km 345/5-6”. a) Effective Span b) Width of bridge c) Clear Span* d) Overall length of bridge

Last Answer : c) Clear Span*

Height gauges should be located at a minimum distance of….. from gate post. (a) 5 m (b) 7.5m (c) 8m* (d)10m

Description : Height gauges should be located at a minimum distance of….. from gate post. (a) 5 m (b) 7.5m (c) 8m* (d)10m

Last Answer : (c) 8m*

Two pumps M and N can fill a cistern in 10m and 15m respectively. If both the pumps are opened simultaneously, after how much time should N be closed so that the cistern is full in 8 minutes ? A) 5min B) 6min C) 3min D) 7min

Description : Two pumps M and N can fill a cistern in 10m and 15m respectively. If both the pumps are opened simultaneously, after how much time should N be closed so that the cistern is full in 8 minutes ? A) 5min B) 6min C) 3min D) 7min

Last Answer :  C X(1/10+1/15) +(8-x)1/10 = 1 5x/30+8-x/10 =1 5x+24-3x/30 =1 2x+24 = 30 2x=6 X=6/2 = 3

Out of HCI and CH3COOH, which one is a weak acid and why? Describe an activity to support your answer. -Chemistry

Description : Out of HCI and CH3COOH, which one is a weak acid and why? Describe an activity to support your answer. -Chemistry

Last Answer : Acetic acid ( CH3COOH) is a weaker acid because it does not dissociate completely into its ions in aqueous solution. Activity: Add zinc metal in HCI and CH3COOH respectively. The hydrogen gas will be evolved ... and light red in CH3COOH which shows HCI is a strong acid and CH3COOH is a weak acid.

What is the full form of 'HCI' ? -How To ?

Description : What is the full form of 'HCI' ? -How To ?

Last Answer : The full form of 'HCI' is Hotel Corporation of India

Já hci měsícčky! :(

Description : I'm 13 and I don't have months. My friends already have it, one is 12 and I got it today. I'm under a lot of pressure. I'd really like them. stupid when classmates talk about it.

Last Answer : Just calm down ... all the local women will confirm to you that there is nothing to stand for ... you just aren't ready yet .... and then they go overgrown and they wouldn't like them best ... except for a few ... it artificially it's a hormonal thing ....: D so dockej casu ... ☢ ☠ ☣ ☠ ☢ ☠ ☣ ☠ ☢

When 22.4 litres of `H_(2)(g)` is mixed with 11.2 litres of `CI_(2)(g)`, eacg at STP, the moles of HCI(g) formed is equal to :-

Description : When 22.4 litres of `H_(2)(g)` is mixed with 11.2 litres of `CI_(2)(g)`, eacg at STP, the moles of HCI(g) formed is equal to :-

Last Answer : When 22.4 litres of `H_(2)(g)` is mixed with 11.2 litres of `CI_(2)(g)`, eacg at STP, the moles of HCI(g ... ) C. 1 mol of HCI (g) D. 2 mol of HCI (g)

`CH_(3)CH_(2)CH(OH)CH(CH_(3))_(2)+CH_(3)COCI overset("base")rarr CH_(3)CH_(2)CH(OCOCH_(3))CH(CH_(3))_(2)+HCI` In the above reaction, if the reactant a

Description : `CH_(3)CH_(2)CH(OH)CH(CH_(3))_(2)+CH_(3)COCI overset("base")rarr CH_(3)CH_(2)CH(OCOCH_(3))CH(CH_(3))_(2)+HCI` In the above reaction, if the reactant a

Last Answer : `CH_(3)CH_(2)CH(OH)CH(CH_(3))_(2)+CH_(3)COCI overset("base")rarr CH_(3)CH_(2) ... the same configuration at the chiral atom D. be optically inactive

Optically active amine having molecular formula `C_(5)H_(13)N` on reaction with `NaNO_(2)+HCI` produces, `3^(@)` optically inative alcohol. Find out s

Description : Optically active amine having molecular formula `C_(5)H_(13)N` on reaction with `NaNO_(2)+HCI` produces, `3^(@)` optically inative alcohol. Find out s

Last Answer : Optically active amine having molecular formula `C_(5)H_(13)N` on reaction with `NaNO_(2)+HCI` produces, ` ... out structures of amines: A. B. C. D.

Which of the following products are formed when 1-propanamine is treated with `NaNO_(2)+HCI`?

Description : Which of the following products are formed when 1-propanamine is treated with `NaNO_(2)+HCI`?

Last Answer : Which of the following products are formed when 1-propanamine is treated with `NaNO_(2)+HCI`? A. B. C. D.