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3 years ago

5.80 gg of NaCl in 350. mL of a NaClsolution

Chemistry

1 answer:

Delicious77 [7]3 years ago

6 0

Answer:

Solution is 0.28 M

You can also say, [NaCl] = 0.28 mol/L

Explanation:

As you have a solute mass and the solution's volume, you may find the molarity concentration of solution.

Molarity specifies the moles of solute in 1 L of solution

We convert the volume of solution to L → 350 mL . 1L / 1000 mL = 0.350L

We convert the mass of solute to moles → 5.80 g . 1mol / 58.45 g = 0.0992 moles

Molarity (mol/L) = 0.0992 mol /0.350L = 0.28M

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Answer:

Explanation:

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3 years ago

How did you get the 3.2?

Helga [31]

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The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

MAVERICK [17]

Answer:

Approximately $2.46\; \rm mol$ .

Explanation:

Make use of the molar mass data ( $M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}$ ) to calculate the number of moles of molecules in that $64.0\; \rm g$ of $\rm C_2H_2$ :

$\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}$ .

Make sure that the equation for this reaction is balanced.

Coefficient of $\rm C_2H_2$ in this equation: $2$ .

Coefficient of $\rm H_2O$ in this equation: $2$ .

In other words, for every two moles of $\rm C_2H_2$ that this reaction consumes, two moles of $\rm H_2O$ would be produced.

Equivalently, for every mole of $\rm C_2H_2$ that this reaction consumes, one mole of $\rm H_2O$ would be produced.

Hence the ratio: $\displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1$ .

Apply this ratio to find the number of moles of $\rm H_2O$ that this reaction would have produced:

$\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}$ .

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3 years ago

The equilibrium constant for the dissolution of silver chloride (AgCl(s) Ag+(aq) + Cl–(aq)) has a value of 1.79 × 10–10. Which s

ryzh [129]

"Silver chloride is essentially insoluble in water" this statement is true for the equilibrium constant for the dissolution of silver chloride.

Option: b

<u>Explanation</u>:

As silver chloride is essentially insoluble in water but also show sparing solubility, its reason is explained through Fajan's rule. Therefore when AgCl added in water, equilibrium take place between undissolved and dissolved ions. While solubility product constant $\left(\boldsymbol{K}_{s p}\right)$ for silver chloride is determined by equilibrium concentrations of dissolved ions. But solubility may vary also at different temperatures. Complete solubility is possible in ammonia solution as it form stable complex as water is not good ligand for Ag+.

To calculate $\left(\boldsymbol{K}_{s p}\right)$ firstly molarity of ions are needed to be found with formula: $\text { Molarity of ions }=\frac{\text { number of moles of solute }}{\text { Volume of solution in litres }}$

Then at equilibrium cations and anions concentration is considered same hence:

$\left[\mathbf{A} \mathbf{g}^{+}\right]=[\mathbf{C} \mathbf{I}]=\text { molarity of ions }$

Hence from above data $\left(\boldsymbol{K}_{s p}\right)$ can be calculated by: $\left(\boldsymbol{K}_{s p}\right)$ = $\left[\mathbf{A} \mathbf{g}^{+}\right] \cdot[\mathbf{C} \mathbf{I}]$

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3 years ago

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erma4kov [3.2K]

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