All categories

4 years ago

Volume of 0.150 M NaOH solution required to neutralize 25mL of a 0.055 M HCl solution?

Chemistry

1 answer:

Inessa05 [86]4 years ago

5 0

<h3>Answer:</h3>

9.17 mL

<h3>Explanation:</h3>

In this question<u> we are given;</u>

Molarity of NaOH as 0.150 M
Molarity of the acid, HCl as 0.055 M
Volume of the acid, HCl as 25 mL
Equation for the reaction as;

NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)

We are required to determine the volume of the alkali, NaOH used;

<h3>Step 1: Determine the moles of the acid, HCl</h3>

Molarity = Number of moles ÷ Volume

Thus, rearranging the formula;

Moles = Molarity × volume

= 0.055 M × 0.025 L

= 0.001375 moles

<h3>Step 2: Moles of the alkali, NaOH</h3>

From the reaction; 1 mole of HCl reacts with 1 mole of NaOH

Therefore;

Moles of HCl = Moles of NaOH

Hence; Moles of NaOH = 0.001375 moles

<h3>Step 3: Determine the volume of the alkali, NaOH</h3>

From the previous equation;

Molarity = Moles ÷ Volume

Rearranging the equation;

Volume = Moles ÷ Molarity

Therefore;

Volume of NaOH = 0.001375 moles ÷ 0.150 M

= 0.00917 L, but, 1 L = 1000 mL

= 9.17 mL

Therefore, the volume of NaOH required is 9.17 mL

You might be interested in

If the mole fraction of nitric acid (HNO3) in an aqueous solution is 0.275, what is the percent by mass of HNO3?

jeka94

Molar mass of N = 14 g/molMolar mass of O2 = 32 g/molAdding both masses = 46 g/molActual molar mass/ Empirical molar mass = 138.02 / 46 = 3Now multiplying this co effecient with empirical fomula NO2 = 3(NO2) = N3O6So according to above explanation,D) N3O6, is the correct answer.

8 0

3 years ago

If the human body were a car, glucose would be:

NeTakaya

<span>If the human body were a car, glucose would be the gasoline.
Glucose gives humans energy, we basically run on glucose, among other things, the same way a car would run on gas.
</span>

8 0

3 years ago

10 points

Murljashka [212]

Answer:

2. The metal would lose one electrons and the non metal would gain one electrons

Explanation:

An atom of a certain element reacts with the atoms of other elements in order to fullfill its outermost shell (called valence shell).

We notice the following:

- The elements in Group 1 (which are metals) have only 1 electron in their valence shell

- The elements in Group 17 (which are non-metals) have 1 vacancy (lack of electron) in their valence shell

This means that in order for both an atom of group 1 and an atom of group 17 to fullfill the valence shell, they have to:

- The atom in group 1 has to give away its only electron of the valence shell

- The atom in group 17 has to gain one electron in order to fullfill the shell

Therefore, the correct option is

2. The metal would lose one electrons and the non metal would gain one electrons

7 0

4 years ago

3) In the Hydrolysis of Disaccharides and Polysaccharides portion of the lab, starch should give one positive iodine test and on

AfilCa [17]

Answer: potassium iodide is the basic test for starch,and the positive test is blue-black coloration, any other test substance which is not starch will give a negative results.

Explanation:

Starch is an example of polysaccharide and since the standard test for it is potassium iodide solution, it gives a positive test.

Diasaccharides e.g maltose are reducing sugars.their standard test is BENEDICT test .

Therefore, in the hydrolysis; starch should give a positve test, while Diasaccharides should give negative rest.

3 0

3 years ago

What mass of butane in grams is necessary to produce 1.5×103 kJ1.5×103 kJ of heat? What mass of CO2CO2 is produced? Assume the r

saul85 [17]

32.8 g of Butane is required and 99.3 g of CO₂ is produced

<u>Explanation:</u>

The above mentioned reaction can be written as,

C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g) where ΔH (rxn)= -2658 kJ

It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.

That is

$$\frac{1500}{2658}=0.564 \text { moles }$ of butane reacted

Now this moles is converted into mass by multiplying it with its molar mass = 0.564 mol × 58.122 g / mol

= 32.8 g of butane.

Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol

= 99.3 g of CO₂

Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced

7 0

3 years ago