When was Carbon discovered? (The element from the periodic table)

Making solutions is an extremely important component to real-life chemistry. If you make 3.00 L of a solution using 90.0 g of so

kumpel [21]

Answer:

Final concentration of NaOH = 0.75 M

Explanation:

For $NaOH$ :-

Given mass = 90.0 g

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{90.0\ g}{39.997\ g/mol}$

$Moles\ of\ NaOH= 2.2502\ mol$

Molarity is defined as the number of moles present in one liter of the solution. It is basically the ratio of the moles of the solute to the liters of the solution.

The expression for the molarity, according to its definition is shown below as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Where, Volume must be in Liter.

It is denoted by M.

Given, Volume = 3.00 L

So,

$Molarity=\frac{2.2502\ mol}{3.00\ L}=0.75\ M$

<u>Final concentration of NaOH = 0.75 M</u>

5 0

4 years ago

An aqueous solution containing 17.5 g of an unknown molecular (nonelectrolyte) compound in 100.0 g of water has a freezing point

Sonbull [250]

Answer:

molar mass = 180.833 g/mol

Explanation:

mass sln = mass solute + mass solvent

∴ solute: unknown molecular (nonelectrolyte)

∴ solvent: water

∴ mass solute = 17.5 g

∴ mass solvent = 100.0 g = 0.1 Kg

⇒ mass sln = 117.5 g

freezing point:

ΔTc = - Kc×m

∴ ΔTc = -1.8 °C

∴ Kc H2O = 1.86 °C.Kg/mol

∴ m: molality (mol solute/Kg solvent)

⇒ m = ( - 1.8 °C)/( - 1.86 °C.Kg/mol)

⇒ m = 0.9677 mol solute/Kg solvent

molar mass (Mw) [=] g/mol

∴ mol solute = ( m )×(Kg solvent)

⇒ mol solute = ( 0.9677 mol/Kg) × ( 0.100 Kg H2O )

⇒ mol solute = 0.09677 mol

⇒ Mw solute = ( 17.5 g ) / ( 0.09677 mol )

⇒ Mw solute = 180.833 g/mol

6 0

3 years ago

Which of the following is a balanced chemical equation?

True [87]

Answer:

equation number 3 is balanced.

hope it helps ☺️!

8 0

3 years ago

Hi! ❤️ , im looking for some help here. ill give brainliest if able to.

dexar [7]

A student builds a model of a race car. The scale is 1:15. In the scale model, the car is 8 cm tall. How tall is the actual car?

<h2>Answers:</h2>

#CarryOnLearning

7 0

2 years ago

II. Ionic Equations

mario62 [17]

Answer:

Complete ionic: $\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Net ionic: $\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

Explanation:

Start by identifying species that exist as ions. In general, such species include:

Soluble salts.
Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ . These three salts will exist as ions:

Each $\rm AgNO_3\, (aq)$ formula unit will exist as one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion.
Each $\rm CaCl_2$ formula unit will exist as one $\rm Ca^{2+}$ ion and two $\rm Cl^{-}$ ions (note the subscript in the formula $\rm CaCl_2\!$ .)
Each $\rm Ca(NO_3)_2$ formula unit will exist as one $\rm Ca^{2+}$ and two $\rm {NO_3}^{-}$ ions.

On the other hand, $\rm AgCl$ is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each $\rm AgNO_3\, (aq)$ formula unit will exist as only one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion. However, because the coefficient of $\rm AgNO_3\, (aq)\!$ in the original equation is two, $\!\rm AgNO_3\, (aq)$ alone should correspond to two $\rm Ag^{+}\!$ ions and two $\rm {NO_3}^{-}\!$ ions.

Do not rewrite the salt $\rm AgCl$ because it is insoluble.

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of $\rm Ca^{2+}$ and two units of $\rm {NO_3}^{-}$ . Doing so will give:

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}$ .

Simplify the coefficients:

$\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

7 0

3 years ago