La lluvia acida es un fenomeno a nivel local o Regional?

Suppose of barium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of barium cati

ollegr [7]

Explanation:

Let us assume that the given data is as follows.

mass of barium acetate = 2.19 g

volume = 150 ml = 0.150 L (as 1 L = 1000 ml)

concentration of the aqueous solution = 0.10 M

Therefore, the reaction equation will be as follows.

$Ba(C_{2}H_{3}O_{2})_{2} \rightarrow Ba^{2+} + 2C_{2}H_{3}O^{-}_{2}$

Hence, moles of $C_{2}H_{3}O^{-}_{2}$ = $2 \times Ba(C_{2}H_{3}O_{2})_{2}$ .......... (1)

As, No. of moles = $\frac{mass}{\text{molar mass}}$

Hence, moles of $Ba(C_{2}H_{3}O_{2})_{2}$ will be calculated as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{2.19 g}{255.415 g/mol}$ (molar mass of $Ba(C_{2}H_{3}O_{2})_{2}$ is 255.415 g/mol)

= $8.57 \times 10^{-3}$

Moles of $C_{2}H_{3}O^{-}_{2}$ = $2 \times 8.57 \times 10^{-3}$

= 0.01715 mol

Hence, final molarity will be as follows.

Molarity = $\frac{\text{no. of moles}}{volume}$

= $\frac{0.01715 mol}{0.150 L}$

= 0.114 M

Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.

5 0

3 years ago

Vitamin K is involved in normal blood clotting. When 0.802 g of vitamin K is dissolved in 25.0 g of camphor, the freezing point

xenn [34]

Answer:

Molar mass of vitamin K = 450.56\frac{g}{mol}[/tex]

Explanation:

The freezing point of camphor = 178.4 ⁰C

the Kf of camphor = 37.7°C/m

where : m = molality

the relation between freezing point depression and molality is

Depression in freezing point = Kf X molality

Where

Kf = cryoscopic constant of camphor

molality = moles of solute dissolved per kg of solvent.

putting values

2.69°C = 37.7°C/m X molality

molality = 0.0714 mol /kg

$molality=\frac{molesofvitaminK}{massofcamphor(kg)}=\frac{moles}{0.025}$

moles of vitamin K = 0.0714X0.025 = 0.00178 mol

we know that moles are related to mass and molar mass of a substance as:

$moles=\frac{mass}{molarmass}$

For vitamin K the mass is given = 0.802 grams

therefore molar mass = $\frac{mass}{moles}=\frac{0.802}{0.00178}=450.56\frac{g}{mol}$

7 0

3 years ago

How many grams of sodium sulfide can be produced when 45.3 g Na react with 105 g S?

erastova [34]

Balance the equation: 2Na + S --> Na2S
Using the given amount of the reactants in the reaction, calculate the amount of the product:
45.3g Na x (1 mol/22.99 g)= 1.97 mol of Na
105f S x (1 mol/ 32.06g) = 3.28 mol of S
The limiting reactant would be Na:
<span>1.97 mol Na x (1 mol Na2S/ 2 mol Na) x (78.04g/mol) = 76.87g of Na2S produced</span>

8 0

3 years ago

Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is ________ atm. (a

svetoff [14.1K]

Answer:

The answer to your question is P = 1.357 atm

Explanation:

Data

Volume = 22.4 L

1 mol

temperature = 100°C

a = 0.211 L² atm

b = 0.0171 L/mol

R = 0.082 atmL/mol°K

Convert temperature to °K

Temperature = 100 + 273

= 373°K

Formula

$(P + \frac{a}{v^{2}} )(v - b) = RT$

Substitution

$(P + \frac{0.211}{22.4})(22.4 - 0.0171) = (0.082)(373)$

Simplify

(P + 0.0094)(22.3829) = 30.586

Solve for P

P + 0.0094 = $\frac{30.586}{22.3829}$

P + 0.0094 = 1.366

P = 1.336 - 0.0094

P = 1.357 atm

7 0

3 years ago

Effect of Two-Step Homogenization on the Evolution of Al3Zr Dispersoids in Al-0.3Mg-0.4Si-0.2Zr Alloy

AlekseyPX

Effect of Two-Step Homogenization on the Evolution of Al3Zr Dispersoids in Al-0.3Mg-0.4Si-0.2Zr Alloy Al3Zr nano-particles can be introduced in Al-Mg-Si 6xxx alloys to improve their elevated temperature behavior and recrystallization resistance. The effect of two-step homogenization treatments on

the precipitation of Al3Zr dispersoids in Al-0.3Mg-0.4Si-0.2Zr alloy was investigated and compared to

<h3>What is Homogenization?</h3>

Any of a number of methods, including homogenization and homogenisation, are used to uniformly combine two liquids that are insoluble in one another. To do this, one of the liquids is changed into a state in which very minute particles are evenly dispersed across the other liquid. The process of homogenizing milk, in which the milk fat globules are equally distributed throughout the remaining milk and reduced in size, is a classic example. In order to create an emulsion, two immiscible liquids (i.e., liquids that are not soluble in all amounts one in another) must be homogenized (from "homogeneous"; Greek, homos, same + genos, kind)[2] (Mixture of two or more liquids that are generally immiscible).

To learn more about Homogenization from the given link:

brainly.com/question/18271118

#SPJ4

7 0

1 year ago