Perform the following

a student adds 3.5 moles of solute to enough water to make a 1500mL solution. what is the concentration?

aksik [14]

<h2>Hello!</h2>

The answer is:

$MolarConcentration=\frac{3.5moles}{volume(1.5L)}=2.33molar$

Since there is not information about the solute but only its mass, we need to assume that we are calculating the molar concentration of a solution or molarity. So, need to use the following formula:

$MolarConcentration=\frac{mass(solute)}{volume(solution)}$

Now, we know that the mass of the solute is equal 3.5 moles and the volume is equal to 1500 mL or 1.5L

Then, substituting into the equation, we have:

$MolarConcentration=\frac{3.5moles}{1.5L}=2.33molar$

Have a nice day!

7 0

2 years ago

Read 2 more answers

Which of the following is consistent with a spontaneous process in the forward direction?

Kryger [21]

Answer:

A) ∆Suniv >0, ∆G<0, T∆Suniv >0.

Explanation:

The connection between entropy and the spontaneity of a reaction is expressed by the <u>second law of thermodynamics</u><u>: The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process</u>.

Mathematically, we can express the second law of thermodynamics as follows:

For a spontaneous process: ΔSuniv = ΔSsys + ΔSsurr > 0

Therefore, the second law of thermodynamics tells us that a spontaneous reaction increases the entropy of the universe; that is, ΔSuniv > 0.

If we want spontaneity expressed only in terms of the properties of the system (ΔHsys and ΔSsys), we use the following equation:

-TΔSuniv = ΔHsys - TΔSsys < 0

That means that T∆Suniv >0.

This equation says that for a process carried out at constant pressure and temperature T, if the changes in enthalpy and entropy of the system are such that <u>ΔHsys - TΔSsys is less than zero, the process must be spontaneous.</u>

Finally, if the change in free energy is less than zero (ΔG<0), the reaction is spontaneous in the forward direction.

7 0

2 years ago

When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier t

lisov135 [29]

Answer:

1.59mol/L

Explanation:

Data obtained from the question include:

Mass of MgCl2 = 151g

Volume of water(solvent) = 1L

Now, let us calculate the number of mole of MgCl2. This is illustrated below:

Molarity Mass of MgCl2 = 24 + (2x35.5) = 24 + 71 = 95g/mol

Mass of MgCl2 = 151g

Number of mole of MgCl2 =?

Number of mole = Mass /Molar Mass

Number of mole of MgCl2 = 151/95

Number of mole of MgCl2 = 1.59mole

Now we can calculate the molarity of MgCl2 as follow:

Mole = 1.59mole

Volume = 1L

Molarity =?

Molarity = mole /Volume

Molarity = 1.59/1

Molarity = 1.59mol/L

8 0

1 year ago

Element M reacts with oxygen to form an oxide with the formula MO. When MO is dissolved in water, the resulting solution is basi

Tpy6a [65]

Answer:

Calcium

Explanation:

Since the element reacts with oxygen to form an oxide with the formula MO, the charge on the element is +2.

Also, since the oxide MO when dissolved in water is basic, the metal is an alkali earth metal.

From the above conditions;

The metal is not arsenic because arsenic is a metalloid has the following oxides As₂O₃ and As₃O₅ and are respectively amphoteric and acidic in nature

The metal is not germanium because is a metalloid and even though germanium oxide has the formula GeO₂, it is amphoteric.

The metal is not chlorine because chlorine is a non-metal

The metal is definitely calcium because calcium oxide has the formula CaO and calcium is an alkaline earth metal.

The metal is not selenium because selenium is anon-meal and its oxide has the formula Se0₂ and is acidic

5 0

2 years ago

A compound is made up of 28 g N, 24 g C, 48 g O, and 8 g H .What is the empirical formula?

vovikov84 [41]

Answer:

$\rm C_2H_8N_2O_3$ .

Explanation:

<h3>Step One: calculate the coefficients. </h3>

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

$\rm C$ : approximately $12$ .
$\rm H$ : approximately $1$ .
$\rm N$ : approximately $14$ .
$\rm O$ : approximately $16$ .

The relative atomic mass of an element is numerically equal to the mass (in grams, $\rm g$ ,) of one mole of atoms of this element.

For example, the relative atomic mass of $\rm C$ is approximately $12$ . Therefore, each mole of $\rm C\!$ atoms would have a mass of $12\; \rm g$ .

This sample contains $24\; \rm g$ of carbon. That would correspond to approximately $\displaystyle \left(\frac{24}{12}\right)\; \rm mol = 2\; \rm mol$ of $\rm C$ atoms.

Similarly, for the other three elements:

$\displaystyle n(\mathrm{H}) \approx \frac{8\; \rm g}{1\; \rm g \cdot mol^{-1}} = 8\; \rm mol$ .

$\displaystyle n(\mathrm{N}) \approx \frac{28\; \rm g}{14\; \rm g \cdot mol^{-1}} = 2\; \rm mol$ .

$\displaystyle n(\mathrm{O}) \approx \frac{48\; \rm g}{16\; \rm g \cdot mol^{-1}} = 3\; \rm mol$ .

Hence, the ratio between these elements in this compound would be:

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ .

In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ is indeed the smallest possible integer ratio between the number of atoms of these elements.

<h3>Step Two: arrange the elements in an appropriate order</h3>

Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:

If carbon, $\rm C$ , is present in this compound, then:

$\rm C$ (carbon) and then $\rm H$ (hydrogen) will be the first two elements listed in the formula (ignore the hydrogen if it is not in the compound.)
The other elements in this compound will be listed in alphabetical order.

If there is no carbon $\rm C$ in this compound, then list all the elements in this compound in alphabetical order.

Both $\rm C$ (carbon) and $\rm H$ (hydrogen) are found in this compound. Therefore, the first element in the list would be $\rm C\!$ . The second would be $\rm H\!$ , followed by $\rm N\!$ and then $\rm O\!$ .

Hence, the empirical formula of this compound would be $\rm C_2H_8N_2O_3$ .

6 0

2 years ago