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3 years ago

C12H7Cl3FNaO2 what are the elements in that chemical formula

Chemistry

1 answer:

Bess [88]3 years ago

3 0

Answer:

hope it helped you

Explanation:

Sodium fluoride - 5-chloro-2-(2,4-dichlorophenoxy)phenol (1:1:1) | C12H7Cl3FNaO2 |

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Use the image to answer the question.

Roman55 [17]

Answer:

A1+3

Explanation:

A1+3 is the most common ion of aluminum.

3 0

3 years ago

The solubility product of calcium fluoride (CaF2(s); fluorite) is 310-11 at 25C. Could a fluoride concentration of 1.0 mg L-1

Finger [1]

The given question is incomplete. The complete question is as follows.

The solubility product of calcium fluoride () is at 25 degrees C. Will a fluoride concentration of 1.0 mg/L be soluble in a water containing 200 mg/L of calcium?

Explanation:

Reaction equation for the given chemical reaction is as follows.

$CaF_{2} \rightleftharpoons Ca^{2+} + 2F^{-}$

Therefore, expression for $K_{sp}$ will be as follows.

$K_{sp} = [Ca^{2+}][F^{-}]^{2}$

Also, moles of per liter = \frac{\text{mass of F^{-} per L}}{\text{molar mass of F}}[/tex]

= $\frac{1.0 \times 10^{-3}}{19.0}$

= $5.263 \times 10^{-5} mol$

Hence, $[F^{-}] = \frac{\text{moles of F^{-}}{volume}$

= $\frac{5.263 \times 10^{-5}}{1}$

= M

Now, moles of per L = \frac{\text{mass of Ca^{2+} per L}}{\text{molar mass of Ca}}[/tex]

= $\frac{200 \times 10^{-3}}{40.1}$

= M

Also, $[Ca^{2+}] = \frac{moles of Ca^{2+}}{volume}$

= $\frac{4.988 \times 10^{-3}}{1}$

= M

Hence, ionic product =

= $(4.988 \times 10^{-3}) \times (5.263 \times 10^{-5})^{2}$

= $1.38 \times 10^{-11}$

As, the ionic product is less than the $K_{sp}$ , this means that the fluoride will be soluble in water containing the calcium.

4 0

3 years ago

Calculate the amount in grams of Na2CO3 needed to react with HCL to produce 120g NaCl

worty [1.4K]

Let's start off with the balanced chemical equation:

$Na_2CO_{3_{(s)}} + \textbf2HCl_{_{(aq)}}\rightarrow CO_{2_{(g)}} + H_2O_{_{(l)}} + \textbf2NaCl_{_{(aq)}}$

Assuming the sodium carbonate is the limiting reagent, look at the coefficients of sodium carbonate and sodium chloride and use those as a ratio of sodium carbonate to sodium chloride: 1:2.

Since you have the required mass of NaCl, convert this to moles.

Assuming you know how to find the molar mass of NaCl:

$M_{NaCl} = 58.44g/mol$
$n = \frac{m}{M}$
$n_{NaCl} = \frac{120g}{58.44g/mol}$
$n_{NaCl} = 2.053mol$

Using the ratio, since 1 mole of sodium carbonate is required to produce 2 moles of sodium chloride, cross-multiply the ratios:

$1:2 = x:2.053mol$
$2x = 2.053mol$
$x = 1.027mol$

Therefore 1.027 moles of sodium carbonate is required to produce the required amount of sodium chloride. Convert to mass for your final answer:

$n_{Na_2CO_3} = 1.027mol$
$M_{Na_2CO_3} = 105.99g/mol$
$m = nM$
$m = (1.027mol)(105.99g/mol)$
$m_{Na_2CO_3} = 108.85g$

Hope this helps :)

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3 years ago

How do humans recharge their brains

liraira [26]

Sleeping? what are the choices

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3 years ago

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How does silicon shape our technological reality?

Georgia [21]

Silicon is the most common element used in electronic devices. They are commonly termed as silicon wafers when incorporated into these electronics. Its beneficial property for this use is being a semi-conductor. It can be combined with other conductors through a process called doping, to make it ideal for making transistors that could amplify electric signals.

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3 years ago