What volume of 6.00 M hydrochloric acid is needed to prepare 500 mL of 0.100 M solution?

galina1969 [7]

Answer:
Molarity = 2.3 M

Explanation:
Molarity can be calculated using the following rule:
Molarity = number of moles of solute / volume of solution

1- getting the number of moles:
We are given that:
mass of solute = 105.96 grams
From the periodic table:
atomic mass of carbon = 12 grams
atomic mass of hydrogen = 1 gram
atomic mass of oxygen = 16 grams
Therefore:
molar mass of C2H6O = 2(12) + 6(1) + 16 = 46 grams
Now, we can get the number of moles as follows:
number of moles = mass / molar mass = 105.96 / 46 = 2.3 moles

2- The volume of solution is given = 1 liter

3- getting the molarity:
molarity = number of moles of solute / volume of solution
molarity = 2.3 / 1
molarity = 2.3 M

Hope this helps :)

8 0

3 years ago

If you weighed out 0.38 g of calcium and reacted it with an excess of hcl, how many moles of h2(g) would you expect to produce?

Hitman42 [59]

Answer is: 0,0095 mol of hydrogen gas will be produced in reaction.
Chemical reaction: Ca + 2HCl → CaCl₂ + H₂.
m(Ca) = 0,38 g.
n(H₂) = ?
n(Ca) = m(Ca) ÷ M(Ca).
n(Ca) = 0,38 g ÷ 40 g/mol
n(Ca) = 0,0095 mol.
from reaction: n(Ca) : n(H₂) = 1 : 1.
n(H₂) = n(Ca) = 0,0095 mol.
n - amount of substance.

4 0

3 years ago

A gold coin contains 3.47 × 10^23<br> gold atoms.<br> What is the mass of the coin in grams?

VashaNatasha [74]

Answer:

Mass = 114.26 g

Explanation:

Given data:

Number of gold atoms = 3.47×10²³ atoms

Mass in gram = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

3.47×10²³ atoms × 1 mol /6.022 × 10²³ atoms

0.58 mol

Mass of gold:

Mass = number of moles × molar mass

Mass = 0.58 mol × 197 g/mol

Mass = 114.26 g

4 0

3 years ago

combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon

masya89 [10]

Answer:

$\rm CH_2$ .

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be $\rm CO_2$ and $\rm H_2O$ . The $\rm C$ and $\rm H$ would come from the hydrocarbon, while the $\rm O$ atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

$\rm C$ : $12.011$ .
$\rm H$ : $1.008$ .
$\rm O$ : $\rm 15.999$ .

Calculate the molar mass of $\rm CO_2$ and $\rm H_2O$ :

$M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$

Calculate the number of moles of $\rm CO_2$ molecules in $33.01\; \rm g$ of $\rm CO_2\!$ :

$\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol$ .

Similarly, calculate the number of moles of $\rm H_2O$ molecules in $13.51\; \rm g$ of $\rm H_2O\!$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol$ .

Note that there is one carbon atom in every $\rm CO_2$ molecule. Approximately $0.7501\; \rm mol$ of $\rm CO_2\!$ molecules would correspond to the same number of $\rm C$ atoms. That is: $n(\mathrm{C}) \approx 0.7501\; \rm mol$ .

On the other hand, there are two hydrogen atoms in every $\rm H_2O$ molecule. approximately $0.7499\; \rm mol$ of $\rm H_2O$ molecules would correspond to twice as many $\rm H\!$ atoms. That is: $n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol$ .

The ratio between the two is: $n(\mathrm{C}): n(\mathrm{H}) \approx 1:2$ .

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be $\rm CH_2$ .

6 0

3 years ago

Does Monocercomonoides still meet the definition of a eukaryote? Why or why not?

KengaRu [80]

Answer:

Yes

Explanation:

They are a unique type of eukaryote because they lack an important organelle: mitochondria. Mitochondria are essential for producing cellular energy in most eukaryotic cells. However, due to its habitat, it is able to acquire energy from a process called sulfur mobilization.

They are significant because they challenge the idea that eukaryotes need mitochondria to be classified as eukaryotic. However, they have other membrane-bound organelles such as a nucleus and Golgi apparatus, meaning they remain eukaryotic.

Research suggest they lost their mitochondria over time, rather than never having had them throughout their ancestry.

Because of all these reasons, they still meet the definition of a eukaryote.

6 0

3 years ago