What is 3623÷72×2(59)1(4)<8335×(3)(2x)

Potassium iodide reacts with lead(II) nitrate in this precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) Wha

andrew11 [14]

Answer:

a. 174 mL

Explanation:

Let's consider the following reaction.

2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)

We have 155.0 mL of a 0.112 M lead(II) nitrate solution. The moles of Pb(NO₃)₂ are:

0.1550 L × 0.112 mol/L = 0.0174 mol

The molar ratio of KI to Pb(NO₃)₂ is 2:1. The moles of KI are:

2 × 0.0174 mol = 0.0348 mol

The volume of a 0.200 M KI solution that contains 0.0348 moles is:

0.0348 mol × (1 L / 0.200 mol) = 0.174 L = 174 mL

5 0

3 years ago

If the lab technician needs 30 liters of a 25% acid solution, how many liters of the 10% and the 30% acid solutions should she m

Mice21 [21]

Answer:

7.5 L of the 10% and 22.5 L of the 30% acid solution, she should mix.

Explanation:

Let the volume of 10% acid solution used to make the mixture = x L

So, the volume of 30% acid solution used to make the mixture = y L

Total volume of the mixture = x + y = 30 L .................. (1)

For 10% acid solution:

C₁ = 10% , V₁ = x L

For 30% acid solution :

C₂ = 30% , V₂ = y L

For the resultant solution of sulfuric acid:

C₃ = 25% , V₃ = 30 L

Using

C₁V₁ + C₂V₂ = C₃V₃

10×x + 30×y = 25×30

So,

x + 3y = 75 .................. (2)

Solving 1 and 2 we get,

x = 7.5 L

y = 22.5 L

6 0

3 years ago

A 215-g sample of copper metal at some temperature is added to 26.6 g of water. The initial water temperature is 22.22 oC, and t

andrezito [222]

The initial temperature of the copper metal was 27.38 degrees.

Explanation:

Data given:

mass of the copper metal sample = 215 gram

mass of water = 26.6 grams

Initial temperature of water = 22.22 Degrees

Final temperature of water = 24.44 degrees

Specific heat capacity of water = 0.385 J/g°C

initial temperature of copper material , Ti=?

specific heat capacity of water = 4.186 joule/gram °C

from the principle of:

heat lost = heat gained

heat gained by water is given by:

q water = mcΔT

Putting the values in the equation:

qwater = 26.6 x 4.186 x (2.22)

qwater = 247.19 J

qcopper = 215 x 0.385 x (Ti-24.4)

= 82.77Ti - 2019.71

Now heat lost by metal = heat gained by water

82.77Ti - 2019.71 = 247.19

Ti = 27.38 degrees

8 0

3 years ago

Quick question plz . .....Which one of the following is acidic oxide?

dedylja [7]

Answer:

B

Explanation:

its an acidic oxide, it disolves in water to form carbonic acid which is an acid

8 0

3 years ago

Consider the balanced equation of K I KI reacting with P b ( N O 3 ) 2 Pb(NOX3)X2 to form a precipitate. 2 K I ( a q ) + P b ( N

professor190 [17]

Answer: 50.7 grams

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $KI$

$\text{Number of moles}=molarity\times {\text {Volume in L]}=0.417M\times 0.528L=0.220moles$

The balanced chemical equation is:

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)$

$KI$ is the limiting reagent as it limits the formation of product and $Pb(NO_3)_2$ is in excess.

According to stoichiometry :

2 moles of $KI$ give = 1 mole of $PbI_2$

Thus 0.220 moles of $KI$ give= $\frac{1}{2}\times 0.220=0.110moles$ of $PbI_2$

Mass of $PbI_2=moles\times {\text {Molar mass}}=0.110moles\times 461.01g/mol=50.7g$

Thus 50.7 g of $PbI_2$ will be formed.

6 0

4 years ago