Aqueous solutions of sodium hypoch lorite (NaOCI), best known as bleach, are prepared by the reaction of sodium hydroxide with chlorine: 2 NaOH (aq)Cl2(g)->NaOCI (aq)+ H20 (I)+ NaCl (aq) How many grams of NaOH are needed to react with 25.0 g of chlorine?

Answers

Answer 1
Answer:

Answer:

28.2 g of NaOH

Explanation:

We need to calculate the grams of NaOH needed to react with 25.0 g of Cl₂ in the following reaction:

2 NaOH(aq) + Cl₂(g) → NaOCI(aq + H₂0(I) + NaCl(aq)

We are going to solve this by making use of the molar ratio between Cl₂ and NaOH given by the reaction equation where we see that every mol of Cl₂ will react with 2 moles of NaOH.

So first we need to convert the 25.0 g of Cl₂ to moles:

  • Molar Mass of Cl₂ = 2 x 35.45 = 70.90 g/mol
  • Moles of Cl₂ = 25.0 g / 70.90 g/mol = 0.3526 moles

Then we need to calculate the moles of NaOH needed to react with these moles of Cl₂ knowing that every mol of Cl₂ will react with 2 moles of NaOH:

  • moles of NaOH = 2 x moles of Cl₂ = 2 x 0.3526 moles = 0.7052 moles

Next we must convert these moles to grams:

  • Molar Mass of NaOH = 22.990 + 15.999 + 1.008 = 40.00 g/mol
  • Mass of NaOH = 0.7052 moles x 40.00 g/mol = 28.2 g

28.2 g are needed to react with 25.0 g of Cl₂ in the production of NaOCl


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If a large marshmallow has a volume of 2.75 in3 and density of 0.242 g/cm3, how much would it weigh in grams? 1 in3=16.39 cm3.

Answers


•   
Volume of the marshmallow:

V = 2.75 in^3          (but, 1 in^3 = 16.39 cm^3)

V = 2.75 × 16.39 cm^3

V = 2.75 × 16.39 cm^3

V = 45.0725 cm^3


•   Density:

d = 0.242 g/cm^3


•  Mass:

m = d × V

m = (0.242 g/cm^3) × (45.0725 cm^3)

m = (0.242 g/cm^3) × (45.0725 cm^3)

m = 10.907545 g

m ≈ 10.9 g   <———  this is the answer.


I hope this helps. =)

Answer:

9.92g

Explanation:

2.50 in31×16.39 cm31 in3×0.242 gcm3=9.92 g

Isn't this false? For the industrial production of indigo carmine, a blue food colouring additive, a synthetic process with an E-factor of 17.4 produces less waste than a synthetic process with an E-factor of 3.0.

The answer I got was False, is this correct?

Answers

Answer: yes it is false

Explanation:

The statement is false. A synthetic process with a lower E-factor produces less waste than a process with a higher E-factor.

The E-factor is a measure of the waste generated during a manufacturing process. It is calculated by dividing the total mass of waste produced by the mass of the desired product. A lower E-factor indicates that less waste is generated per unit of product.

In this case, the synthetic process with an E-factor of 3.0 produces less waste than the process with an E-factor of 17.4. This means that the process with an E-factor of 3.0 is more efficient in terms of waste reduction.

Arrange the following aqueous solutions in order of decreasing freezing points: 0.10 m KNO3, 0.10 m BaCl2, 0.10 m ethylene glycol ​[C2H4(OH)2, antifreeze], and 0.10 m Na3PO4.

Answers

Answer:

The freezing point of a solution is lowered compared to the freezing point of the pure solvent. The amount of depression of the freezing point is proportional to the molality of the solute. The greater the molality of a solution, the lower its freezing point. To compare the freezing points of these solutions, we need to determine which one has the highest molality.

First, we need to determine the number of particles that each solute will produce in solution, as this affects the amount of depression of the freezing point.

KNO3 dissociates into two ions: K+ and NO3-, so it will produce two particles per formula unit.

BaCl2 dissociates into three ions: Ba2+ and two Cl-, so it will produce three particles per formula unit.

Ethylene glycol does not dissociate in solution, so it will produce one particle per molecule.

Na3PO4 dissociates into four ions: three Na+ and one PO43-, so it will produce four particles per formula unit.

Now, we can calculate the molality (moles of solute per kilogram of solvent) for each solution:

For 0.10 m KNO3: molality = 0.10 mol / 1 kg = 0.10 m

For 0.10 m BaCl2: molality = 0.10 mol x 3 particles / 1 kg = 0.30 m

For 0.10 m ethylene glycol: molality = 0.10 mol / 1 kg = 0.10 m

For 0.10 m Na3PO4: molality = 0.10 mol x 4 particles / 1 kg = 0.40 m

So, the solutions in order of decreasing freezing points are:

0.10 m Na3PO4 (highest molality)

0.10 m BaCl2

0.10 m KNO3 and 0.10 m ethylene glycol (same molality, but KNO3 has a smaller van't Hoff factor than ethylene glycol, so it will have a slightly higher freezing point)

Explanation:

How many sp2 and sp3 are in caffeine?

Answers

Answer:

5.56×10²² molecules of AlPO4

Explanation:

i think thats is the correct answer if its wrong im really sorry

Final answer:

Caffeine has two sp2 and one sp3 hybridized carbon atoms in its structure.

Explanation:

Caffeine has three carbon atoms in its structure, and the hybridization of these carbon atoms determines the types of bonds they form. Two of the carbon atoms in caffeine are sp2-hybridized, while the remaining carbon atom is sp3-hybridized.

The sp2-hybridized carbon atoms form double bonds with nitrogen atoms and are found in the pyrimidine and pyrazole rings of the caffeine molecule. They have trigonal planar geometry with bond angles of approximately 120 degrees.

The sp3-hybridized carbon atom is found in the methyl group (CH3) attached to the pyrimidine ring. It forms single bonds with three hydrogen atoms and has tetrahedral geometry with bond angles approximately 109.5 degrees.

Learn more about Chemical Bonding here:

brainly.com/question/33579397

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A biochemist studying breakdown of the insecticide DDT finds that it decomposes by a first-order reaction with a half-life of 12.0 yr. How long does it take DDT in a soil sample to decompose from 809 ppbm to 10.0 ppbm (parts per billion by mass)

Answers

Answer:

76.0 years

Explanation:

Step 1: Given data

  • Half-life (t1/2): 12.0 y
  • Initial concentration of DDT ([A]₀): 809 ppbm
  • Final concentration of DDT ([A]): 10.0 ppbm
  • Time elapsed (t): ?

Step 2: Calculate the rate constant (k)

We will use the following expression.

k = ln 2/ t1/2 = ln 2 / 12.0 y = 0.0578 y⁻¹

Step 3: Calculate the time elapsed

For a first-reaction order, we will use the following equation.

ln [A]/[A]₀ = -k × t

t = ln [A]/[A]₀ / (-k)

t = (ln 10.0 ppbm/809 ppbm) / (-0.0578 y⁻¹)

t = 76.0 y

What is the symbol of the element that is classified as an alkali metal and is in period 4?

Answers

Answer:

Potassium (K) [First element in period 4]