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Lina20 [59]
3 years ago
10

0.250 moles of NaCl is dissolved in 0.850 L of solution. What is the molar concentration?

Chemistry
1 answer:
Nuetrik [128]3 years ago
6 0

Considering the definition of molarity, the molar concentration is 0.294 \frac{moles}{liter}.

Molarity reflects the concentration of a solution indicating the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

molarity=\frac{amount of moles of solute}{volume}

Molarity is expressed in units \frac{moles}{liter}.

In this case, you know:

  • amount of moles of solute= 0.250 moles
  • volume= 0.850 L

Replacing in the definition of molarity:

molarity=\frac{0.250 moles}{0.850 L}

Solving:

molarity= 0.294 \frac{moles}{liter}

Finally, the molar concentration is 0.294 \frac{moles}{liter}.

Learn more about molarity with this example: brainly.com/question/15406534?referrer=searchResults

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Carbon monoxide (CO) gas reacts with oxygen (O2) gas to produce carbon dioxide (CO2) gas. If 1.00 L of carbon monoxide reacts wi
Lunna [17]
1) Chemical equation:

2CO + O2 ---> 2CO2

2) molar ratios 2 moles CO : 1 mol O2 : 2 moles CO2.

3) When temperature and pressure is kept constant, the molar ratios are equal to the volume ratios.

So, at the same temperature and pressure conditions (standard) you ca state

2 L CO : 1 LO2 : 2 L CO2

=> 2L CO : 2 L CO2 => 1L CO : 1 L CO2.

So, 1 liter of CO2 is produced when 1 liter of CO reacts with excess O2.
7 0
3 years ago
Read 2 more answers
Calcium Carbonate reacts with dilute hydrochloric acid. The equation for the reaction is shown. CaCo3 + 2Hcl = Cacl2 + H2O + Co2
Elena-2011 [213]

Answer:

Approximately 0.224\;\rm L, assuming that this reaction took place under standard temperature and pressure, and that \rm CO_2 behaves like an ideal gas. Also assume that the reaction went to completion.

Explanation:

The first step is to find out: which species is the limiting reactant?

Assume that \rm CaCO_3 is the limiting reactant. How many moles of \rm CO_2 would be produced?

Look up the relative atomic mass of \rm Ca, \rm C, and \rm O on a modern periodic table:

  • \rm Ca: 40.078.
  • \rm C: 12.011.
  • \rm O: 15.999.

Calculate the formula mass of \rm CaCO_3:

\begin{aligned} & M(\rm CaCO_3) \\ &= 40.078 + 12.011 + 3 \times 15.999 \\&= 100.086\; \rm g \cdot mol^{-1}\end{aligned}.

Calculate the number of moles of formula units in 1\; \rm g of \rm CaCO_3 using its formula mass:

\begin{aligned}& n(\mathrm{CaCO_3})\\&= \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} \\ &= \frac{1\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \approx 1.00\times 10^{-2}\; \rm mol\end{aligned}.

In the balanced chemical equation, the ratio between the coefficient of \rm CaCO_3 and that of \rm CO_2 is \displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{CaCO_3})} = 1.

In other words, for each mole of \rm CaCO_3 formula units consumed, one mole of \rm CO_2 would be produced.

If \rm CaCO_3 is indeed the limiting reactant, all that approximately 1.00\times 10^{-2}\; \rm mol of \rm CaCO_3\! formula would be consumed. That would produce approximately 1.00\times 10^{-2}\; \rm mol\! of \rm CO_2.

On the other hand, assume that \rm HCl is the limiting reactant.

Convert the volume of \rm HCl to \rm dm^{3} (so as to match the unit of concentration.)

\begin{aligned}&V(\mathrm{HCl})\\ &= 50\; \rm cm^{3} \\ &= 50\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{10^{3}\; \rm cm^{3}} \\ &= 5.00\times 10^{-2}\; \rm dm^{3} \end{aligned}.

Calculate the number of moles of \rm HCl molecules in that 5.00\times 10^{-2}\; \rm dm^{3} of this \rm 0.05\; \rm mol \cdot dm^{-3}

\begin{aligned}& n(\mathrm{HCl}) \\ &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.05\; \rm mol \cdot dm^{-3}\\ &\quad\quad \times 5.00\times 10^{-2}\;\rm dm^{3} \\ &= 2.50 \times 10^{-3}\; \rm mol\end{aligned}.

Notice that in the balanced chemical reaction, the ratio between the coefficient of \rm HCl and that of \rm CO_2 is \displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{HCl})} = \frac{1}{2}.

In other words, each mole of \rm HCl molecules consumed would produce only 0.5\;\rm mol of \rm CO_2 molecules.

Therefore, if \rm HCl is the limiting reactant, that 2.50 \times 10^{-3}\; \rm mol of \rm HCl\! molecules would produce only one-half as many (that is, 1.25\times 10^{-3}\; \rm mol) of \rm CO_2 molecules.

If \rm CaCO_3 is the limiting reactant, \rm 1.00\times 10^{-3}\; \rm mol of \rm CO_2 molecules would be produced. However, if \rm HCl is the limiting reactant, 1.25\times 10^{-3}\; \rm mol of \rm CO_2\! molecules would be produced.

In reality, no more than \rm 1.00\times 10^{-3}\; \rm mol of \rm CO_2 molecules would be produced. The reason is that all \rm CaCO_3 would have been consumed before \rm HCl was.

After finding the limiting reactant, approximate the volume of the \rm CO_2\! produced.

Assume that this reaction took place under standard temperature and pressure (STP.) Under STP, the volume of one mole of ideal gas molecules would be approximately 22.4\; \rm L.

If \rm CO_2 behaves like an ideal gas, the volume of that \rm 1.00\times 10^{-3}\; \rm mol of \rm CO_2\! molecules would be approximately \rm 1.00\times 10^{-3}\; \rm mol \times 22.4\; \rm L = 0.224\; \rm L.

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3 years ago
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Explanation :

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This reaction is a redox reaction or oxidation-reduction reaction in which sulfur get oxidized and oxygen get reduced.

(c) 2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)

This reaction is a combination reaction in which the two reactants molecule combine to form a large molecule or product.

(d) 2AlN\rightarrow 2Al(s)+N_2(g)

This reaction is a decomposition reaction in which a large molecule or reactant decomposes to give two or more molecule or products.

(e) BaCl_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaCl(aq)

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This reaction is a combination reaction in which the two reactants combine to form a large molecule or product.

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This reaction is a double displacement reaction in which the cation of two reactants molecule exchange their places to give two different products.

(h) 2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(l)

This reaction is combustion reaction in which a hydrocarbon react with an oxygen to give carbon dioxide and water as a products.

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