Calculate the number of milliliters of 0.587 M NaOH required to precipitate all of the Fe3 ions in 197 mL of 0.654 M FeCl3 solution as Fe(OH)3. The equation for the reaction is: FeCl3(aq) 3NaOH(aq) Fe(OH)3(s) 3NaCl(aq)

Answers

Answer 1

Answer: The number of milliliters of 654 mL for 0.587 M NaOH required to precipitate all of the [tex]Fe^{3+}[/tex] ions in 197 mL of 0.654 M [tex]FeCl_{3}[/tex] solution as [tex]Fe(OH)_{3}[/tex].

Explanation:

The reaction equation is as follows.

[tex]FeCl_{3}(aq) + 3NaOH(aq) \rightarrow Fe(OH)_{3}(s) + 3NaCl(aq)[/tex]

Therefore, moles of [tex]Fe(OH)_{3}[/tex] are calculated as follows.

Moles = Molarity of [tex]Fe(OH)_{3}[/tex]  [tex]\times[/tex] Volume (in L)

= 0.654 M [tex]\times[/tex] 0.197 L  

= 0.128 mol

Now, according to the given balanced equation 1 mole of [tex]FeCl_{3}(aq)[/tex] reacts with 3 moles of NaOH(aq). Hence, moles of [tex]Fe(OH)_{3}[/tex]  reacted are calculated as follows.

3 [tex]\times[/tex] 0.128 mol = 0.384 moles of NaOH

As moles of NaOH present are as follows.

Moles of NaOH = Molarity of NaOH [tex]\times[/tex] Volume (in L)

0.384 mol = 0.587 M [tex]\times[/tex] Volume (in L)

Volume (in L) = 0.654 L (1 L = 1000 mL) = 654 mL

Thus, we can conclude that the number of milliliters of 654 mL for 0.587 M NaOH required to precipitate all of the [tex]Fe^{3+}[/tex] ions in 197 mL of 0.654 M [tex]FeCl_{3}[/tex] solution as [tex]Fe(OH)_{3}[/tex].


Related Questions

Calculate the amount of heat required to completely sublime 55.0 g of solid dry ice CO2 at its sublimation temperature. The heat of sublimation for carbon dioxide is 32.3 kj mol

Answers

Answer:

40.4 kJ

Explanation:

Step 1: Given data

Mass of CO₂ (m): 55.0 gHeat of sublimation of CO₂ (ΔH°sub): 32.3 kJ/mol

Step 2: Calculate the moles corresponding to 55.0 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

n = 55.0 g × 1 mol/44.01 g = 1.25 mol

Step 3: Calculate the heat (Q) required to sublimate 1.25 moles of CO₂

We will use the following expression.

Q = n × ΔH°sub

Q = 1.25 mol × 32.3 kJ/mol = 40.4 kJ

Nitric acid (HNO3) reacts with ammonia (NH3) in aqueous solution. Use your knowledge of nitric acid to decide what type of reaction arrow(s) to use. $$ Part 2 (1 point) Sulfuric acid (H2SO4) reacts with ammonia in aqueous solution. Use your knowledge of sulfuric acid to decide what type of reaction arrow(s) to use. $$

Answers

Answer:

Both reactions are acid-base reactions

Explanation:

An acid base reaction is a reaction that occurs between an acid and a base. This reaction often leads to the formation of a salt in the process. The nature of the salt depends on the type of acid and base that reacted in the process.

Both HNO3 and H2SO4 are strong acids. However, ammonia is a weak base. The acid base reaction between ammonia and these strong acids is shown below;

HNO3(aq) + NH3(aq) ------>NH4NO3(aq)

H2SO4(aq) + 2NH3(aq) ----> (NH4)2SO4(aq)

A student sets up the following equation to convert a measurement.
(The stands for a number the student is going to calculate.)

Fill in the missing part of this equation.

Answers

Answer:

–0.13 Pa.m²

Explanation:

From the question given above, the following data were obtained:

Measurement (Pa.mm²) = –1.3×10⁵ Pa.mm²

Measurement (Pa.m²) =?

We can convert from Pa.mm² to Pa.m² by doing the following:

1 Pa.mm² = 1×10¯⁶ Pa.m²

Therefore,

–1.3×10⁵ Pa.mm² = –1.3×10⁵ Pa.mm² × 1×10¯⁶ Pa.m² / 1 Pa.mm²

–1.3×10⁵ Pa.mm² = –0.13 Pa.m²

Thus, –1.3×10⁵ Pa.mm² is equivalent to –0.13 Pa.m².

The complete equation will be:

[tex](-1.3\times 10^5 Pa.mm^2)\times 10^{-6}=(-0.13) Pa.m^2[/tex]

Explanation:

Given:

The equation to convert a measurement:

[tex](-1.3\times 10^5 Pa.mm^2)\times ? = ? Pa.m^2[/tex]

To find:

The missing part of the equation.

Solution:

[tex](-1.3\times 10^5 Pa.mm^2)\times ? = ? Pa.m^2[/tex]

On LHS the unit is in [tex]Pa. mm^2[/tex] and RHS the unit is in [tex]Pa.m^2[/tex] which means that we have to convert [tex]mm^2[/tex] to [tex]m^2[/tex]

In 1 millimeter there are 0.001 meters.

[tex]1 mm = 0.001 m\\1 mm^2=0.000001 m^2=10^{-6} m^2[/tex]

So, the complete equation will be:

[tex](-1.3\times 10^5 Pa.mm^2)\times 10^{-6}=(-0.13) Pa.m^2[/tex]

Learn more about conversions here:

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The value of ΔH° for the reaction below is -6535 kJ. ________ kJ of heat are released in the combustion of 16.0 g of C6H6 (l)?

2C6H6 (l) + 15O2 (g) → 12CO2 (g) + 6H2O (l)

Answers

Answer:

the value of H° is below -6535 kj. +6H2O

Explanation:

6H2O answer solved

For the given reaction, 2 moles of C₆H₆ the heat energy released is - 6535 KJ. Then, for 16 g of the compound or 0.205 moles needs 669.83 KJ of heat released in combustion.

What is combustion ?

Combustion is a chemical reaction that occurs between a fuel and an oxidizing agent, typically oxygen, resulting in the release of heat, light, and various combustion products, such as carbon dioxide and water vapor.

The process of combustion involves a rapid and exothermic (heat-releasing) oxidation reaction that produces a flame, which is visible in many cases.

Here, 2 moles of the hydrocarbon releases - 6535 KJ of energy.

molar mass of C₆H₆ = 78 g/mol

then no.of moles in 16 g = 16 /78 = 0.205 moles.

then energy released by 0.205 moles  = 0.205 moles × 6535 KJ /2 moles = 669.83 kJ

Therefore, the heat energy released by 16 g of the compound in combustion is 669.83 kJ.

Find more on combustion :

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#SPJ2

how old was the oldest animal fossil
help thx​

Answers

Answer:

the Rhyniognatha hirsti

Explanation:

at age 400 million years old

In the presence of excess iodide ions, the iodine formed by reaction of iodide with NBS will react further to form triiodide ions. What does the triiodide combine with to form the blue color of the endpoint

Answers

Answer:

Starch.

Explanation:

When the triiodide combine with starch, it forms dark blue colour. Amylose in starch is responsible for the occurrence of a deep blue color when the iodine is combine with the starch. The iodine molecule goes inside of the amylose coil which makes a linear triiodide ion complex that goes into the coil of the starch that leads to an intense blue-black color in the end so we can say that starch turns the colour into blue.

Acetylide ions react with aldehydes and ketones to give alcohol addition products.

a. True
b. False

Answers

Answer:

a

Explanation:

Star
Planet
*
As the planet makes one completer revolution around the star, starting at the position shown the gravitational attraction between the star
and the planet will
A Continually decrease
3 Decrease, then increase
increase then decrease
Romain the same
RI
12.20 AM
618/2001

Answers

Answer:

according to the path shown in the figure it will start decreasing then again it will start increasing when the path will be nearer to the star.

Reason is gravitation force is indirectly proportional to the distance.

So, option B. decrease then increase is correct

38. Consider the following equilibrium:
2CO(g) + O2(g) =2CO2
Keg=4.0 x 10-10
What is the value of Key for 2CO2(g) + 2COR + O2g) ?​

Answers

Answer:

[tex]Key=2.5x10^{-9}[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the equilibrium constant value for the reverse reaction:

[tex]2CO_2(g) \rightleftharpoons 2CO(g) + O_2(g)[/tex]

By knowing that the equilibrium expression is actually:

[tex]Key =\frac{[CO]^2[O_2]}{[CO_2]^2} =\frac{1}{Keg}[/tex]

Thus, we plug in and solve for the inverse of Keq to obtain Key as follows:

[tex]Key =\frac{1}{4.0x10^{-10}}\\\\Key=2.5x10^{-9}[/tex]

Regards!

In the reaction represented by the equation: N2 + 3H2 → 2NH3, what is the conversion factor of nitrogen to ammonia? Explain by using law of definite proportion

Answers

Answer:

10/3

Explanation:

Hope this helps

Answer:

The compound formula for the ammonia is

N

H

3

. It is a colorless gas. It is the result of the chemical reaction between nitrogen and hydrogen gas. The chemical reaction between the gases is shown below:

N

2

+

3

H

2

2

N

H

3

If 4.00 moles of O2 occupies a volume of 5.0 L at a particular temperature and pressure, what volume will 3.00 moles of oxygen gas occupy under the same condition?

Answers

Answer: Volume occupied by 3.00 moles of oxygen gas under the same condition is 3.75 L.

Explanation:

Given: [tex]n_{1}[/tex] = 4.00 moles,          [tex]V_{1}[/tex] = 5.0 L

[tex]n_{2}[/tex] = 3.00 moles,                 [tex]V_{2}[/tex] = ?

Formula used is as follows.

[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}[/tex]

Substitute the values into above formula as follows.

[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{5.0 L}{4.00 mol} = \frac{V_{2}}{3.00 mol}\\V_{2} = 3.75 L[/tex]

Thus, we can conclude that volume occupied by 3.00 moles of oxygen gas under the same condition is 3.75 L.

If we have 1.23 mol of NaOH in solution and 0.85 mol of Cl2 gas is available to react, which one is the limiting reactant? Give your reason.​

Answers

Answer:

NaOH is the limiting reactant.

Explanation:

Hello there!

In this case, since the reaction taking place between sodium hydroxide and chlorine has is:

[tex]NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]

Which must be balanced according to the law of conservation of mass:

[tex]2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]

Whereas there is a 2:1 mole ratio of NaOH to Cl2, which means that the moles of the former that are consumed by 0.85 moles of the latter are:

[tex]n_{NaOH}=0.85molCl_2*\frac{2molNaOH}{1molCl_2}\\\\n_{ NaOH}=1.7molNaOH[/tex]

Therefore, since we just have 1.23 moles out of 1.70 moles of NaOH, we infer this is the limiting reactant.

Regards!

someone answer please​

Answers

uhhh maybe decreases

Answer:

A

Explanation:

It took 2.30 minutes using a current of 3.00 A to plate out all the copper from 0.300 L of a solution containing Cu2 . What was the original concentration of Cu2

Answers

Answer:

7.16 × 10⁻³ M

Explanation:

Let's consider the reduction reaction of copper during the electroplating.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We can calculate the moles of Cu²⁺ present in the solution using the following relations.

1 A = 1 C/s.1 min = 60 s.1 mole of electrons has a charge of 96486 C (Faraday's constant).1 mole of Cu²⁺ is reduced when 2 moles of electrons are gained.

The moles of Cu²⁺ reduced are:

[tex]2.30 min \times \frac{60s}{1min} \times \frac{3.00C}{s} \times \frac{1mole^{-} }{96486C} \times \frac{1molCu^{2+} }{2mole^{-} } = 2.15 \times 10^{-3} molCu^{2+}[/tex]

[tex]2.15 \times 10^{-3} moles[/tex] of Cu²⁺ are in 0.300 L of solution.

[Cu²⁺] = 2.15 × 10⁻³ mol/0.300 L = 7.16 × 10⁻³ M

5.96 g of ammonia reacts completely according to the following reaction:
2 NH, (g) + Co, (g) → CN,OH, (s) + H20 (1)
(a) What is the theoretical yield of urea (CN,OH,) for this reaction?
(b) If 13.74 g of urea are produced, what is the percent yield for this equation?

please show work, will give brainliest

Answers

Explanation:

this explanation may help u to understand:)

HELPP

There are 9.23 x 1023 molecules of water in a beaker, how many moles are there?

Answers

Answer:

Answer: There are 1.53 moles present in   molecules of water in a beaker.

Explanation:

According to the mole concept, there are  molecules present in 1 mole of a substance.

So, number of moles present in  molecules are calculated as follows.

Thus, we can conclude that there are 1.53 moles present in   molecules of water in a beaker.

Explanation:

cuáles son las características de la luz y en qué consisten​

Answers

Answer:

Cuáles son las características de la luz y en qué consisten?

Explanation:

La luz es una radiación que se propaga en forma de ondas. Las ondas que se pueden propagar en el vacío se llaman ONDAS ELECTROMAGNÉTICAS. La luz es una radiación electromagnética

At a fixed volume, a four-fold increase in the temperature of a gas will lead to _______ in pressure.
Question 2 options:

A)

no change

B)

a two-fold decrease

C)

a four-fold decrease

D)

a four-fold increase

Answers

Answer:

D) a four-fold increase

Explanation:

According to Gay-Lussac's law, which states that the pressure of a given amount of gas is directly proportional to the temperature at a constant volume, the pressure increases with an increase in temperature.

According to this question, at a fixed volume, a four-fold increase in the temperature of a gas will lead to a four-fold increase in the pressure as well.

Chrysanthenone is an unsaturated ketone. If Chrysanthenone has M+ = 150 and contains 2 double bond(s) and 2 ring(s); what is its molecular formula? Enter the formula in the form CH first, then all other atoms in alphabetical order; do not use subscripts. The formula is case-sensitive.

Answers

Answer:

the Molecular formula will be; C10H14O

Explanation:

Given the data in the question;

Chrysanthenone is an unsaturated ketone,

it has M+ = 150 and contains 2 double bond(s) and 2 ring(s).

molecular formula = ?

we know that ketone contain 1 oxygen and mass of oxygen is 16

so mass of the C and H remaining will be;

⇒ 150 - 16 = 134

Now we determine the number of C atoms;

⇒ 134 / 13 = 10

hydrocarbon with 10 hydrogen atom have CnH2n+2 means

⇒ ( 10 × 2 ) +2 = 22 hydrogens

But then we have 3 unsaturation meaning 6 hydrogens less and also we have ring meaning 2 more hydrogens

⇒ 22 - 6 - 2 = 14

Hence the Molecular formula will be; C10H14O

a) If we have a 4.5 L container of CH 10 gas at a temperature of 178 K and a pressure of 0.50 atm, then how many moles of CaHio do
we have?
b) How many grams of C4H1o do we have?

Answers

Answer:

a) 0.15 mol.

b) 8.95 g.

Explanation:

Hello there!

In this case, according to the given information, it is possible for us to infer this problem is solved by using the ideal gas equation:

[tex]PV=nRT[/tex]

And proceed as follows:

a) Here, we solve for the moles, n,  as follows:

[tex]n=\frac{PV}{RT} \\\\n=\frac{0.50atm*4.5L}{0.08206\frac{atm*L}{mol*K}*178K} \\\\n=0.15mol[/tex]

b) for the calculation of the mass, we recall the molar mass of butane, 58.12 g/mol, to obtain:

[tex]0.15mol*\frac{58.12g}{1mol} =8.95g[/tex]

Regards!

20ml of water is mixed with 40gm of fine powder. Calculate the concentration of the solution obtained.

Answers

Answer:

[tex]\%m=66.7\%[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the concentration of the solution obtained, by knowing 20 mL of water are the same to 20 g and therefore the mass of the solution is 40g+20g=60g.

Next, we apply the following equation to obtain the required concentration:

[tex]\%m=\frac{40g}{60g} *100\%\\\\\%m=66.7\%[/tex]

Regards!

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