When energy is transferred from the air to the water, what happens to most of the energy?

Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the fina

yaroslaw [1]

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

Explanation:

n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V(solution) = 150mL = 0.15L

C(AgNO3) = 35mM = 0.035M = 0.035m/L

n(AgNO3) = C(AgNO3) x V(solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI(3) = AgI(3) + FeNO3

So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol

C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M

8 0

3 years ago

Cutting a piece of paper in half proves that solids have no definite shape.<br><br> True or False?

erma4kov [3.2K]

False. Solids have a definite shape and volume. Cutting a paper (solid) in half will alter the shape of the original solid and give it a new shape and volume instead. A solid will always have a definite shape. If not, it's not a solid.

8 0

3 years ago

How many molecules are in 85 grams of Silver?

densk [106]

Answer:

3.0 × 10

$3.0 \times {10}^{23}$

molecules

Explanation:

there are many molecules

8 0

3 years ago

What information allows you to conclude that this article is reliable? Check all that apply.more than 50 studies were conducted.

Hitman42 [59]

Answer:MORE THAN 50 STUDIES WERE CONDUCTED

THE US PUBLIC HEALTH SERVICE WAS RESPONSIBLE FOR PART OF THE STUDY

Explanation: Just completed the assignment Egenuity , i hope this helps .

7 0

3 years ago

Read 2 more answers

This question involves two calculations. The answer to the first part will be

ozzi

Answer :

(a) The mass of $Al_2O_3$ produced is, 15.2 grams.

(b) The percent yield of the reaction is, 72.5 %

Explanation :

Part (a) :

Given,

Mass of $Al$ = 85.1 g

Molar mass of $Al$ = 27 g/mol

First we have to calculate the moles of $Al$

$\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{85.1g}{27g/mol}=3.15mol$

Now we have to calculate the moles of $Al_2O_3$

The balanced chemical equation is:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the reaction, we conclude that

As, 4 moles of $Al$ react to give 2 moles of $Al_2O_3$

So, 3.15 moles of $Al$ react to give $\frac{2}{4}\times 3.15=1.58$ mole of $Al_2O_3$

Now we have to calculate the mass of $Al_2O_3$

$\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3\times \text{ Molar mass of }Al_2O_3$

Molar mass of $Al_2O_3$ = 102 g/mole

$\text{ Mass of }Al_2O_3=(1.58moles)\times (102g/mole)=161.2g$

Therefore, the mass of $Al_2O_3$ produced is, 161.2 grams.

Part (b) :

Now we have to calculate the percent yield of the reaction.

$\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 116.9 g

Theoretical yield = 161.2 g

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

$\text{Percent yield}=\frac{116.9g}{161.2g}\times 100=72.5\%$

Therefore, the percent yield of the reaction is, 72.5 %

8 0

3 years ago