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3 years ago

You need to separate two substances that make up a compund hoe coukld u go about doing it

1 answer:

5 0

Compounds are chemically separated unlike mixtures. To separate one you could burn it or dissolve it in something to make the compound separate into its two elements.

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

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

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago

An archeologist finds a 1.62 kg goblet that she believes to be made of pure gold. She determines that the volume of the goblet i

maxonik [38]

Density (d) which is the quotient when mass (m) is divided by volume (v) is usually reported in terms of g/mL.
d = m /v
Substituting the known values,
d = (1.62 kg) x (1000 g/ 1 kg) / (205 mL)
The answer would be approximately equal to 7.9 g/mL.

3 0

3 years ago

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What is the molality of a solution of water and kcl if the freezing point of the solution is –3mc030-1.jpgc?

Natasha_Volkova [10]

We will use the expression for freezing point depression ∆Tf
∆Tf = i Kf m
Since we know that the freezing point of water is 0 degree Celsius, temperature change ∆Tf is
∆Tf = 0C - (-3°C) = 3°C
and the van't Hoff Factor i is approximately equal to 2 since one molecule of KCl in aqueous solution will produce one K+ ion and one Cl- ion:
KCl → K+ + Cl-
Therefore, the molality m of the solution can be calculated as
3 = 2 * 1.86 * m
m = 3 / (2 * 1.86)
m = 0.80 molal

4 0

3 years ago

Trisha and her lab partner filled a beaker with 150 milliliters of a 0.10 M HCl solution. How many moles of HCl are in the beake

oksian1 [2.3K]

M=mol/liter
We know that we have 150ml=.15 L and .1 mol of HCl
Rearranging the molarity equation, we get
mol=M*l
mol=(.15)(.1)
=.015 mol

7 0

3 years ago

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A chemist wants to make 6.5 L of a .350M CaCl2 solution. What mass of CaCl2(in g) should the chemist use?

kotykmax [81]

First M stands for Molarity which is (moles of solute) / (Liters of solution). we also know that moles = (mass) / (molar mass). so we can form some equations here. We know:
Molarity (M) = moles (mol) / Liters (L)
moles (mol) = (mass) / (molar mass)

we can substitute the (mass) / (molar mass) for (moles) and get:
M = [(mass) / (molar mass)] / Liters

we can now isolate mass and get
M * Liters * molar mass = mass

now we need to find the molar mass of CaCl2 which is 110.98 g/mol

plug the values in and get
.350M * 6.5L * 110.98 g/mol = mass

mass = 252.4795g however the 6.5L has only 2 sig figs so i would say

mass CaCl2 = 2.5 * 10 ^2 g

5 0

3 years ago

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