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MariettaO [177]
3 years ago
12

Calculate the volume in liters of a 0.00231M copper(II) fluoride solution that contains 175.g of copper(II) fluoride CuF2. Be su

re your answer has the correct number of significant digits.
Chemistry
1 answer:
Free_Kalibri [48]3 years ago
7 0

Answer:

Volume = 746 L

Explanation:

Given that:- Mass of copper(II) fluoride = 175 g

Molar mass of copper(II) fluoride = 101.543 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{175\ g}{101.543\ g/mol}

Moles_{copper(II)\ fluoride}= 1.7234\ mol

Also,

Considering:

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

So,,

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

Given, Molarity = 0.00231 M

So,

Volume =\frac{1.7234}{0.00231}\ L

<u>Volume = 746 L</u>

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hat is the percentage of water in the following compound? Answer using three significant figures. Sodium carbonate decahydrate,
BigorU [14]
You need to find the whole molar mass of the compound using the periodic table to add the values. 

Na2CO3= (2 x 23.0) + 12.0 + (3 x 16.0)= 106 g/mol
H2O= 10 x [ (2 x 1.01 ) + (16.0) ]= 180.2 g/mol

the total molar mass is 106 + 180.2 = 286.2 g/mol

the percentage of water you can find by doing "parts over the whole"

H2O%= 180.2 / 286.2 X 100= 63.0%
7 0
3 years ago
To what category of elements does an element belong if it's a poor conductor of electricity
Karo-lina-s [1.5K]

Answer:

I would assume NON-METALS

7 0
4 years ago
If a solution contains 3 moles/liter of sodium chloride (nacl, made of sodium ions and chloride ions), what is the osmolarity of
valkas [14]

<u>Answer:</u> Osmolarity of the sodium chloride solution is 18 Osmol/L.

<u>Explanation:</u>

Osmolarity is defined as the the concentration of the solution which is expressed as the total number of solution particles present in one liter of solvent.

We are given the molarity of the solution which is 3mol/L and to convert it into osmolarity, we will multiply the number of osmoles that are produced by the solute.

Osmole is defined as the particles that contribute to the osmotic pressure of a solution.

The solute given here is sodium chloride (NaCl). Number of osmoles can be determined by the dissociation of this solvent into ions.

The equation given by the dissociation of NaCl:

NaCl\rightarrow Na^++Cl^-

1 mole of sodium chloride produces 2 moles of ions.

So, 3 moles of sodium chloride will produce = (3 × 2) = 6 moles of ions.

Hence, osmolarity of the sodium chloride solution will be 6\times 3mol/L=18Osmol/L

5 0
3 years ago
_Fe2O3 + 2CO —&gt; _Fe + _CO2
soldi70 [24.7K]
<h3>Answer:</h3>

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

<h3>Explanation:</h3>

Concept tested: Balancing of chemical equations

  • A chemical equation is balanced by putting appropriate coefficients on the products and reactants of the equation.
  • Balancing chemical equations ensures that chemical equations obey law of conservation of mass.
  • In this case; to balance the above equation we put the coefficients, 1, 3, 2, and 3 on the reactants and products.
  • Therefore; the balanced chemical equation for the reaction is;

      Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

8 0
4 years ago
A mixture contains NaHCO3 together with unreactive components. A 1.75 g sample of the mixture reacts with HA to produce 0.561 g
Lynna [10]

Answer:

\%NaHCO_3=61.2\%

Explanation:

Hello.

In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:

NaHCO_3+HA\rightarrow NaA+H_2O+CO_2

For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:

m_{NaHCO_3}=0.561gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molNaHCO_3}{1molCO_2} *\frac{84gNaHCO_3}{1molNaHCO_3} \\\\m_{NaHCO_3}=1.071g

Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:

\%NaHCO_3=\frac{1.071g}{1.75g}*100\%\\ \\\%NaHCO_3=61.2\%

Best regards.

5 0
3 years ago
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