Identify 10 different Element's that Exhibit Isotopy.?

A scientist wants to make a solution of tribasic sodium phosphate, na3po4, for a laboratory experiment. How many grams of na3po4

natima [27]

Answer :

The correct answer is for mass of Na₃PO₄ 39.7 g.

Given : 1) Molarity of Na⁺ ions = 1.00 M or 1.00 $\frac{mol}{L}$

2) Volume of solution = 725 mL

Converting volume of solution from mL to L :

Conversion factor : 1 L = 1000 mL

$Volume of solution = 725 mL * \frac{1L }{1000 mL}$

Volume of solution = 0.725 L

Following steps can be done to find mass of :

Step 1 : Write the dissociation reaction of Na₃PO₄ .

$Na_3PO_4 3 Na^+ + PO_4^3^-$

Step 2: Find moles of Na⁺ ions :

Mole of Na⁺ ions can be calculated using molarity formula which is : $Molarity (\frac{mol}{L} ) = \frac{mole of Na^+ }{volume of solution }$

Plugging value of Molarity and volume

$1.00 \frac{mol}{L} = \frac{Mole of Na^+ ions}{0.725 L}$

Multiplying both side by 0.725 L

$1.00 \frac{mol}{L}* 0.725 L = \frac{mole of Na^+ ions}{0.725 L} * 0.725 L$

Mole of Na⁺ ions = 0.725 mol

Step 3: Find mole ratio of Na₃PO₄ : Na⁺ :

Mole ratio is found from coefficients from balanced reaction as:

Mole of Na₃PO₄ in balanced reaction = 1

Mole of Na⁺ ion = 3

Hence mole ratio of Na₃PO₄: Na⁺ = 1 : 3

Step 4 : To find mole of Na₃PO₄

Mole of Na₃PO₄ can be calculated using mole of Na⁺ ion and Mole ratio as :

$Mole of Na_3PO_4= Mole of Na^+ * Mole ratio$

$Mole of Na_3PO_4 = 0.725 mol * \frac{1 mole of Na_3PO_4}{3 mole of Na^+ }$

Mole of Na₃PO₄ = 0.242 mol

Step 5 : To find mass of Na₃PO₄

Mole of Na₃PO₄ can be converted to mass of Na₃PO₄ using molar mass of Na₃PO₄ as :

$Mass (g) = mole (mol) * molar mass \frac{g}{mol}$

$Mass of Na_3PO_4 = 0.242 mol * 163.94 \frac{g}{mol}$

Mass of Na₃PO₄ = 39.619 g

Step 6 : To round off mass of Na₃PO₄ to correct sig fig .

The sig fig in 750 mL and 1.00 M is 3 . So mass of Na₃PO₄ can rounded to 3 sig fig as :

Mass of Na₃PO₄ = 39.7 g

8 0

2 years ago

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

slega [8]

Answer:

The catalyzed reaction will take time of $5.11\times 10^{-8} years$ .

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate of reaction with catalyst

$K_1$ = rate of reaction without catalyst

$Ea_2$ = activation energy with catalyst = 59.0 kJ/mol = 59000 J/mol

$Ea_1$ = activation energy without catalyst = 184 kJ/mol = 184000 J/mol

R = gas constant

T = temperature = $600K$

Now put all the given values in this formula, we get:

$\frac{K_2}{K_1}=e^{\frac{184,000 kJ-59000 kJ}{R\times 300}}=7.632\times 10^{10}$

The reaction enhances by $7.632\times 10^{10}$ when catalyst is present.

Time taken by reaction without catalyzed = 3900 years

Time taken by reaction with catalyzed = x

$x=\frac{3900 year}{7.632\times 10^{10}}=5.11\times 10^{-8} years$

The catalyzed reaction will take time of $5.11\times 10^{-8} years$ .

8 0

2 years ago

Why are biopharmaceuticals in high demand?

elena-s [515]

Answer:

the market is largely driven by the growing population, increasing burden of chronic disease, and rising inclination toward targeted therapy.

Explanation:

also, the huge demand of biopharmaceutical is facilitated by an accelerating focus in research and related investment.

3 0

2 years ago

A gaseous substance turns into a solid. Which best describes this change? A substance that has a specific shape changes to a sub

Snezhnost [94]

A substance that has no specific volume changes to a substance that has a specific volume.

8 0

3 years ago

Read 2 more answers

Which of the following is not a reason why chemical bonds are important

velikii [3]

What are the followings?

4 0

3 years ago

Identify 10 different Element's that Exhibit Isotopy.?​

Identify 10 different Element's that Exhibit Isotopy.?