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Harlamova29_29 [7]
1 year ago
8

you are given a solution of lead nitrate. use any of the solution used in this experiment to form a white precipitate of lead ch

loride. write down the equation
Chemistry
1 answer:
Vesnalui [34]1 year ago
7 0

We have as a reagent a salt, lead nitrate (Pb(NO3)2), and an unknown solution that gives us as a product lead chloride (PbCl2). That is, the solution must contain chlorine.

If a chlorine solution is used we will have the following reaction:

Pb(NO_{3)2}+2Cl^-\rightarrow PbCl_2+2NO^-_3

So, with a chlorine solution, we will have a white precipitate of lead chloride.

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The cloud with those characteristics is #1
5 0
3 years ago
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How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?
xenn [34]

Answer:

5.225x10^{22}atoms\ I

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I

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4 0
3 years ago
If you want to use a serial dilution to make a 1/50 dilution. The first dilution you make is a 1/5 dilution with a total volume
uranmaximum [27]

Answer:

0.2 mL stock solution, 0.8 solvent, 0.1 mL first solution and 0.9 solvent

Explanation:

The final volume for fist solution is 1 mL and concentration must will be 1/5, then 1 mL/5=0.2 mL. For complete the 1 mL add the missing solvent volume 1 mL-0.2 mL=0.8 mL. For second solution, assuming final volume is 1 mL, and concentration 1/10, then we have 1 mL /10=0.1 mL solution 1/5. Completing volume, 1 mL-0.1 mL= 0.9 mL solvent.

7 0
3 years ago
The reaction 2a → a2​​​​​ was experimentally determined to be second order with a rate constant, k, equal to 0.0265 m–1min–1. if
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The integrated rate law for a second-order reaction is given by:

\frac{1}{[A]t} =   \frac{1}{[A]0} + kt

where, [A]t= the concentration of A at time t,

[A]0= the concentration of A at time t=0

<span>k =</span> the rate constant for the reaction


<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min


Hence, \frac{1}{[A]t} = \frac{1}{4} + (0.0265 X 180)

<span>                                        = 4.858</span>

<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>

<span>
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<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M

7 0
3 years ago
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Can anyone help with 5 through 8? Please :(
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Answer:

The answer to your question is below

Explanation:

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4 0
3 years ago
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