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3 years ago

What is the molarity of a HCl solution that

Chemistry

2 answers:

ipn [44]3 years ago

5 0

Answer:

M = 1.04 M

Explanation:

Given data:

Molarity of solution = ?

Mass of HCl = 6.27 g

Volume of solution = 163 mL (163 mL× 1L /1000 mL = 0.163 L)

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Number of moles of HCl:

Number of moles = mass/molar mass

Number of moles = 6.27 g / 36.5 g/mol

Number of moles = 0.17 mol

Molarity:

M = 0.17 mol/ 0.163 L

M = 1.04 M

sergejj [24]3 years ago

3 0

Answer:

4.6 M HCI

Explanation:

on edge

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3Fe + 4H2O Fe3O4 + 4H2

aalyn [17]

These are five questions with its five answers.


First, we have to explain main question.


The statement provides the chemical equation for the reaction of Fe with water to produce iron(III) oxide and hydrogen.


Fe3O4 is a weird chemical formula. It belongs to the product named oxoiron.


Next, I have to tell how you must interpret the question. The five questions are based on the complete reaction of the same number of moles as the coefficients indicated in the chemical equation.


Those coefficients are 3 for Fe, 4 for H₂O, 1 for Fe₃O₄ and 4 for H₂.

With that understood, let's work every question.

1) How many molecules of H₂ are produced?

Answer: 4 moles of molecules.

Justification:


This is, the number of moles of H₂ produced is given by the coefficient indicated in the chemical equation.


2) How many oxygen atoms are required?


Answer: 4.

This is, the atoms of oxygen are supplied in the molecules of water. Since the coeffcient of water is 4, and each molecule o fwater has 1 atom of oxygen, 4 moles of water contain 4 moles of atoms of oxygen.


3) How many moles of Fe₃O₄ are formed?


Answer: 1.


Justification: the coefficient of for formula Fe₃O₄ is 1, indicating that the theoretical yield is 1 mol of molecules.


4) What is the mole ratio of Fe to H₂O?


Answer: 3:4


Justification:


The ratio is the quotient of the two coefficients: the coefficient of the Fe divided by the coefficient of the H₂O.


5) How many hydrogen atoms are involved in this reaction?


Answer: 8 moles of hydrogen atoms.


Justification: as you can see each molecule of H₂O has 2 atoms of hydrogen, then 4 moles of molecules of H₂O have 8 moles of atoms of hydrogen. And of course the same number are in the produt: 4 moles of H₂ contain 8 moles of atomos of hydrogen

3 0

3 years ago

Read 2 more answers

Un tecnico di laboratorio deve preparare una soluzione di carbonato di sodio decaidrato, Na2CO3⋅ 10 H2O per eseguire alcune anal

fiasKO [112]

Answer:

1. 3.70 g Na₂CO₃·10H₂O

2. 50.0 mL of the first solution

Explanation:

1. Prepare the solution

(a) Calculate the molar mass of Na₂CO₃·10H₂O

$\begin{array}{rrr}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 2\times22.99 & 45.98\\\text{1C} & 1\times 12.01 & 12.01\\\text{13O}&13 \times16.00 & 208.00\\\text{20H}&20 \times 1.008 & 20.16\\&\text{TOTAL =} & \mathbf{286.15}\\\end{array}$

The molar mass of Na₂CO₃·10H₂O is 286.15 g/mol.

(b) Calculate the moles of Na₂CO₃·10H₂O

$\text{Moles of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\= \text{0.250 L solution} \times \dfrac{\text{0.0500 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 L solution}}\\\\= \text{0.0125 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}$

(c) Calculate the mass of Na₂CO₃·10H₂O

$\text{Mass of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O }\\= \text{0.012 50 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O } \times \dfrac{\text{296.15 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}\\\\= \text{3.70 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\\text{You need $\large \boxed{\textbf{3.70 g}}$ of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}$

2. Dilute the solution

We can use the dilution formula to calculate the volume needed.

V₁c₁ = V₂c₂

Data:

V₁ = ?; c₁ = 0.0500 mol·L⁻¹

V₂ = 100 mL; c₂ = 0.0250 mol·L⁻¹

Calculation:

$\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\V_{1}\times \text{0.0500 mol/L} & = & \text{100 mL} \times\text{0.0250 mol/L}\\0.0500V_{1}& = & \text{2.500 mL}\\V_{1}&=& \dfrac{\text{2.500 mL}}{0.0500}\\\\& = & \text{50.0 mL}\\\end{array}\\\text{You need $\large \boxed{\textbf{50.0 mL}}$ of the concentrated solution.}$