All categories

2 years ago

If you evaporated 125 mL of a 3.5 M solution of iron(II) nitrite, how many moles of iron(II) nitrite would you recover?

Chemistry

1 answer:

Verizon [17]2 years ago

3 0

Considering the definition of molarity, 0.4375 moles of iron(II) nitrite you would recover.

In order to find the moles of the solid that can be recovered from this solution, you need to use the definition of molarity.

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

$Molarity=\frac{number of moles}{volume}$

Molarity is expressed in units $\frac{moles}{liters}$ .

Moles of iron(II) nitrite

In this case, you evaporated 125 mL of a 3.5 M solution of iron(II) nitrite. Then you know:

Molarity= 3.5 M
number of moles= ?
volume= 125 mL= 0.125 L (being 1000 mL= 1 L)

Replacing in the definition of molarity:

$3.5 M=\frac{number of moles}{0.125 L}$

Solving:

number of moles= 3.5 M× 0.125 L

<u><em>number of moles= 0.4375 moles</em></u>

Finally, 0.4375 moles of iron(II) nitrite you would recover.

Learn more about molarity:

brainly.com/question/9324116

brainly.com/question/10608366

brainly.com/question/7429224

You might be interested in

A nail rusts when the iron in the nail reacts very slowly with oxygen to form a new substance and heat. this is an example of a(

Softa [21]

It's a FUSION reaction because Iron and Oxygen react together to make Iron Oxide (rust) and it's EXOTHERMIC because the reaction forms heat (stated above), which means it's releasing energy.

4 0

3 years ago

Which planet is sometimes called Uranus’s twin? Saturn Pluto Jupiter Neptune

sattari [20]

Neptune.

Because the size, mass, composition and rotation of Neptune and Uranus are very similar they're sometimes refered to as planetary twins.

7 0

3 years ago

Read 2 more answers

The first-order decay of radon has a half-life of 3.823 days. How many grams of radon decomposes after 5.55 days if the sample i

Leni [432]

Answer:

The mass of radon that decompose = 63. 4 g

Explanation:

R.R = P.E/(2ᵇ/ⁿ)

Where R.R = radioactive remain, P.E = parent element, b = Time, n = half life.

Where P.E = 100 g , b = 5.55 days, n = 3.823 days.

∴ R.R = 100/ $2^{5.55/3.823}$

R.R = 100/ $2^{1.45}$

R.R = 100/2.73

R.R = 36.63 g.

The mass of radon that decompose = Initial mass of radon - Remaining mass of radon after radioactivity.

Mass of radon that decompose = 100 - 36.63

= 63.37 ≈ 63.4 g

The mass of radon that decompose = 63. 4 g

8 0

3 years ago

What can be described as the breaking of old bonds and forming of new ones? A. chemical reaction B. atomic state C. conduction D

Neko [114]

Pretty sure its A)chemical reaction

3 0

3 years ago

Which aqueous solution would have the lowest vapor pressure at 25°c 1 M NaCl?

aleksley [76]

The question is incomplete, here is a complete question.

Which aqueous solution would have the lowest vapor pressure at 25°c.

A) 1 M NaCl

B) 1 M $K_3PO_4$

C) 1 M $C_{12}H_{10}O_{11}$

D) 1 M $MgCl_2$

E) 1 M $C_6H_{12}O_6$

Answer : The correct option is, (B) 1 M $K_3PO_4$

Explanation :

According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.

1 M means that the 1 moles of solute present in 1 liter of solution.

Formula used :

$\frac{\Delta p}{p^o}=i\times X_B$

where,

$p^o$ = vapor pressure of the pure component (water)

$p_s$ = vapor pressure of the solution

$X_B$ = mole fraction of solute

i = Van't Hoff factor

As we know that the vapor pressure depends on the mole fraction of solute and the Van't Hoff factor.

So, the greater the number of particles of solute dissolved the lower the resultant vapor pressure.

(a) The dissociation of 1.0 M $NaCl$ will be,

$NaCl\rightarrow Na^++Cl^-$

So, Van't Hoff factor = Number of solute particles = $Na^++Cl^-$ = 1 + 1 = 2

(b) The dissociation of 1 M $K_3PO_4$ will be,

$K_3PO_4\rightarrow 3K^{+}+PO_4^{3-}$

So, Van't Hoff factor = Number of solute particles = $3K^{+}+PO_4^{3-}$ = 3 + 1 = 4

(c) The dissociation of 1 M $C_{12}H_{10}O_{11}$ is not possible because it is a non-electrolyte solute. So, the Van't Hoff factor will be, 1.

(d) The dissociation of 1.0 M $MgCl_2$ will be,

$MgCl_2\rightarrow Mg^{2+}+2Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Mg^{2+}+2Cl^{-}$ = 1 + 2 = 3

(e) The dissociation of 1 M $C_6H_{12}O_{6}$ is not possible because it is a non-electrolyte solute. So, the Van't Hoff factor will be, 1.

From this we conclude that, 1 M $K_3PO_4$ has the highest Van't Hoff factor which means that the solution will exhibit the lowest vapor pressure.

Hence, the correct option is, (B) 1 M $K_3PO_4$

8 0

3 years ago