What type of symbolic relationship do a bat and cow have ?

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

3 years ago

What mass in grams would 5.7L of hydrogen gas occupy at STP?

tekilochka [14]

Answer: The correct answer is: " 0.54 g " .

__________________________________________

Explanation:

Note that "hydrogen gas" is:

H₂ (g) ; that is: a "diatomic element" (diatomic gas) ;

_________________________________________

The molecular weight of "H" is: 1.00794 g ;

(From the Periodic Table of Elements).

So, the molecular weight of: H₂ (g) is:

" 1.00794 g * 2 = 2.01588 g ; {use calculator) ;

_________________________________________

Note the conversion for a gas at STP:

______

1 mol of a gas = 22.4 L gas;

___

i.e. " 1 mol / 22.4 L " ;

____

So: " 5.7 L H₂ (g) $* \frac{1 mol H_{2} }{22.4 L} *\frac{2.01588 g}{mol} =? ;$

The "L" ("literes" cancel out to "1" ; since "L/L = 1 ;

The "mol" (moles) cancel out to "1" ; since "mol/mol = 1 ;

____

and we are left with:

____

[5.7 * 2.104588 g ] / 22.4 = ? g ;

______________________

→ [ 11.9961516 g ] / 22.4 =

0.53554248214 g ;l

_____________________________

We round this value to: " 0.54 g " ;

→ since "5.7 L " has 2 (two) significant figures;

22.4 is an exact number conversion;

and "5.7 L" has fewer significant figures than:

" 2.104588 " ; or: " 1.00794 " .

→ as such: We round to "2 (two) significant figures."

______________________________

Hope this is helpful. Wishing you the best in your academic endeavors!

_______________________________

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Answer:

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Explanation:

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