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Lostsunrise [7]
3 years ago
12

Calculate the molarity of 29.0g of ethanol (C2H5OH) in 545 mL of solution.

Chemistry
2 answers:
raketka [301]3 years ago
7 0
<span>There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. Molarity of the solution given is calculated as follows:

M = (29.0 g ethanol) (1 mol / 46.07 g) / 545mL (1L / 1000mL)
M = 1.16 mol / L </span>
BigorU [14]3 years ago
5 0
Molar mass ethanol:

C2H5OH = 12 x 2+ 1 x 5 + 16  + 1 =  46.0 g/mol

volume = 545 mL in liters: 545 / 1000 => 0.545 L

number of moles:

29.0 / 46.0 => 0.6304 moles

M = n / V

M = 0.6304 / 0.545

M = 1.156 mol/L

hope this helps!


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You need to make an aqueous solution of 0.207 M calcium acetate for an experiment in lab, using a 300 mL volumetric flask. How m
jeka94

Answer:

We have to add 9.82 grams of calcium acetate

Explanation:

Step 1: Data given

Molarity of the calcium acetate solution = 0.207 M

Volume = 300 mL = 0.300 L

Molar mass calcium acetate = 158.17 g/mol

Step 2: Calculate moles calcium acetate

Moles calcium acetate = molarity * volume

Moles calcium acetate = 0.207 M * 0.300 L

Moles calcium acetate = 0.0621 moles

Step 3: Calculate mass calcium acetate

Mass calcium acetate = moles * molar mass

Mass calcium acetate = 0.0621 moles * 158.17 g/mol

Mass calcium acetate = 9.82 grams

We have to add 9.82 grams of calcium acetate

5 0
2 years ago
what is the balanced equation when copper metal is placed in a solution when platnium ii chloride is placed. what is the equatio
koban [17]

Answer:

Cu~+~PtCl_2->Pt~+~CuCl_2

Explanation:

In this case, we can start with the <u>formula of Platinum (II) Chloride</u>. The cation is the atom at the left of the name (in this case Pt^+^2) and the anion is the atom at the right of the name (in this case Cl^-). With this in mind, the <u>formula would be</u> PtCl_2.

Now, if we used <u>metallic copper</u> we have to put in the reaction only the <u>copper atom symbol</u> Cu. So, we have as reagents:

Cu~+~PtCl_2->

The question now is: <u>What would be the products?</u> To answer this, we have to remember <u>"single displacement reactions"</u>. With a general reaction:

A~+~BC->AB~+~C

With this in mind, the reaction would be:

Cu~+~PtCl_2->Pt~+~CuCl_2

I hope it helps!

5 0
2 years ago
In the following reaction, H3PO4 (aq) + H2O (l) ⇄ H2PO4– (aq) + H3O+ (aq) what happens when more H2PO4– (aq) is added to the sol
kirza4 [7]

Answer:

The concentration of H₃PO₄ will increase.

Explanation:

H₃PO₄(aq) + H₂O(l) ⇄ H₂PO₄⁻(aq) + H₃O⁺(aq)

According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

If we add more H₂PO₄⁻, the position of equilibrium will move to the left to get rid of the added H₂PO₄⁻.

The concentration of H₃PO₄ will increase.

7 0
3 years ago
If you combine 27.1 g of a solute that has a molar mass of 27.1 g/mol with 100.0 g of a solvent, what is the molality of the res
baherus [9]

<u>Answer:</u> The molality of the solution is 0.1 m.

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute = 27.1 g

M_{solute} = Molar mass of solute = 27.1 g/mol

W_{solvent} = Mass of solvent = 100 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{27.1\times 1000}{27.1\times 100}\\\\\text{Molality of solution}=0.1m

Hence, the molality of the solution is 0.1 m.

8 0
3 years ago
Read 2 more answers
Please help this is due today! 5-11-2021
Licemer1 [7]

Answer:

- The chemical reaction is not balanced. There is two oxygens on the reactant's side while there's only one oxygen on the products side.

- I would not say it's following the law of conservation of mass as it's not a balanced equation.

- To balance this equation, you would need to add the coefficient of '2' to Magnesium (Mg) on the reactants side, and add the coefficient of '2' to the products side.  This would make it so that there's 2 Mg's and 2 O's on both the reactant's side and products side.

edit: I hope this helped you in some way. ^^

8 0
2 years ago
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