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o-na [289]
3 years ago
11

Determine the concentrations of mgcl2, mg2 , and cl– in a solution prepared by dissolving 2.75 × 10–4 g mgcl2 in 1.75 l of water

. express all three concentrations in molarity. additionally, express the concentrations of the ionic species in parts per million (ppm).
Chemistry
2 answers:
inysia [295]3 years ago
8 0

Answer:


M of MgCl₂ = 1.65 × 10⁻⁶ M


M of Mg²⁺ = 1.65 × 10⁻⁶ M


M of Cl⁻ = 3.30 × 10⁻⁶ M



Explanation:



1) MgCl₂


Molarity = number of moles of solute / volume of solution in liters, M = n / V


n = mass in grams / molar mass


molar mass of MgCl₂ = 24.305 g/mol + 2(35.543 g/mol) = 95.211 g/mol


n = 2.75 × 10⁻⁴ g / 95.211 g/mol = 2.89×10⁻³ moles


⇒ M = n / V = 2.89×10⁻³ moles / 1.75 l = 1.65 × 10⁻⁶ M



2) Mg²⁺ and Cl⁻


Those are the ions in solution.


You assume 100% dissociation of the ionic compound (strong electrolyte).


Then the equation is: MgCl₂ → Mg²⁺ + 2Cl⁻


That means that 1 mol of MgCl₂ produces 1 mol of Mg²⁺ and 2 moles of Cl⁻.


That yields the same molarity concentration of Mg²⁺ , while the molarity concentration of Cl⁻ is the double.



So, the results are:


M of MgCl₂ = 1.65 × 10⁻⁶ M


M of Mg²⁺ = 1.65 × 10⁻⁶ M


M of Cl⁻ = 3.30 × 10⁻⁶ M

Anvisha [2.4K]3 years ago
7 0

Answer 1) : To calculate the concentration of number of moles of MgCl_{2}, by dissolving 2.75 X 10^{-4} of MgCl_{2} will be like this;


Number of moles of MgCl_{2} = (2.75 X 10^{-4} g of MgCl_{2}) X (1 mole of MgCl_{2} / 95.211 g of MgCl_{2})


= 2.88 X 10^{-6} moles of MgCl_{2} (g/L)


In ppm it will be 0.288 ppm


Answer 2) To find the moles of Mg ions in solution of MgCl_{2}, we need to find the moles of Mg present in the solution,


Here, moles of Mg = Moles of MgCl_{2}


So, 2.88 X 10^{-6} moles of MgCl_{2} = 2.88 X 10^{-6} moles of Mg. (g/L)


And in ppm it will be 0.288 ppm


Answer 3) For calculating the moles of Cl ions present in the solution of MgCl_{2}, we need to know how many moles are present in MgCl_{2},


We see 2 moles of Cl ions are present in solution of MgCl_{2}


Therefore, 2 moles of Cl = 1 mole of MgCl_{2}


So, We can multiply 2 with the molarity of MgCl_{2}, we get,


2 X (2.88 X 10^{-6})= 5.77 X 10^{-6} g/L


And in ppm it will be 0.577 ppm.

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How do you solve stoichiometry problems
nataly862011 [7]
Ok so the way I do it is as simple as possible.
Firstly look at the reactants and products ( there can be one reactant and one product or more ) you will usually be given the moles of the reactant or products, if you are given grams you can convert into moles by this convertion ( grams/R.M.M ) where R.M.M is the relative atomic mass of your substance ( the mass number of all of the elements in your substance).

Ok when you have moles now look at the ratio between the products and reactants. Usually you will won't know the moles of one substance therefore you will be asked to find moles or mass of that substance.

For example:

When 16 grams of oxygen and 1 gram of hydrogen gas react to produce water. Find the number of grams of water being produced.

O2 + 2H2 -> 2H2O
16g      2g          xg

Here we're told the mass of the reactants. In stoichiometry we need to work with moles therefore you need to calculate moles of the reactants.
Firstly find the R.M.M of each reactant.
R.M.M of O2 is 16+16=32 since it's diatomic we add atomic masses of two oxygen atoms.
R.M.M of H2 is 1+1=2, it's also diatomic. (Diatomic two atoms of the same element are joined together). (Ignore the number 2 in front of H2, this number shows us the ratio relationship between reactans or products, i.e when we balance an equation.)

Ok so now find moles:

We have 16 grams of O2
16/R.M.M
16/32 = 0.5 moles

We have 2 grams of H2
1/R.M.M
2/2 = 1 mole
Now back to the equation.

O2 + 2H2 -> 2H2O
0.5 moles 1mole xmoles                  (it's xmoles because we don't know                                                               molarity of water that's what we have                                                           find firstly in order to find grams.)
Now look at the ratio between any reactant and product i.e you can choose which reactant to compare to the product, it doesn't make a different ( I will do two or you can do two at the same time)

1st method:
Look at the ratio between O2 and H2O from the reaction above we see the ratio is 1:2 therefore for every 0.5 moles of O2 you get 1 mole of H2O.
1:2
0.5 : x
0.5*2 = 1

2nd method;
Look at the ratio betweem H2 and H2O from the reaction above we see the ratio is 2:2 or 1:1. We have 1 mole of H2 there we must have 1 mole of H2O. We see this is true as both methods give us 1 mole of H2O.

3rd method ( combined):
Look at the ratio between O2, H2 and H2O.
We see that the ratio is 1:2:2
So we have 0.5:1:x
If we multiply 0.5 *2 it equals 1 mole
If we multiply 1*1 we get 1 moles.
Any method is correct and it's up to you to find a comfortable way.
We're not finished in the question we are asked for the mass of water.
So just multiply the number of moles (1mole) by R.M.M of H2O.
1 * R.M.M
R.M.M of H2O = 1+1+16=18
1*18= 18 grams.
And you're finished.

I am sorry if this is so long I want you to understand as much as possible.
In stoichiometry you can also be asked about the empirical formula of a substance. I can show you how do it. If you have any question just tell me.
Hope this helps :).
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3 years ago
Given the elements chlorine, iodine, oxygen, bromine, and fluorine, organize by increasing atomic size (atomic radius). Justify
astra-53 [7]

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3 0
2 years ago
What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o
Nina [5.8K]

Answer:

The concentration of NaCl = 0.3374 M

Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

1 mL= 10^{-3} L

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

The formula can be written for the calculation of moles as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Thus,  

Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}

Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles

Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles

The chemical reaction taking place:

AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)

According to reaction stoichiometry:

<u>1 mole</u> of AgNO₃ reacts with <u>1 mole</u> of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

1 mL= 10^{-3} L

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}

Molarity\ of\ NaCl= 0.3374 M

<u>Thus, the concentration of NaCl = 0.3374 M</u>

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Which substance is the reducing agent in this reaction? 16h++2cr2o72−+c2h5oh→4cr3++11h2o+2co2 express your answer as a chemical
lana66690 [7]

16 H + + 2Cr2O72- + C2H5OH → 4 Cr3 + +11H2O +2CO2

The reducing agent is C2H5OH

Explanation

reducing  agent is a substance that loses or donate electrons  in a chemical reaction. C2H5OH is the one which donate electrons in the above chemical equation.

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