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3 years ago

H3PO4 nomenclatura quimica

Chemistry

1 answer:

svetoff [14.1K]3 years ago

4 0

La nomenclatura química de H3PO4 es Ácido fosfórico

Or in English
Phosphoric acid

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If the air in a room has lower temperature than the air coming out of a floor vent from a forced air heater , heat would flow

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To the top. Common knowledge my dude. Heat always rises to the top.

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3 years ago

Read 2 more answers

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

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

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

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2 years ago

A lab worker receives an average radiation

cricket20 [7]

Answer:

0.0504

Explanation:

4 0

2 years ago

Calculate the pH after the addition of 10.0 mL of 0.240 M sodium hydroxide to 50.0 mL of 0.120 M acetic acid.

Goryan [66]

Answer:

pH = 4.58

Explanation:

The reaction of NaOH with acetic acid, CH₃COOH occurs as follows:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

<em>Moles that react:</em>

NaOH = 10mL = 0.010L * (0,240mol / L) = 0.0024 moles NaOH

CH₃COOH = 50.0mL = 0.050L * (0.120mol / L) = 0.0060 moles CH₃COOH

That means after the reaction you will have:

CH₃COOH: 0.0060 mol - 0.0024 mol = 0.0036 moles

CH₃COO⁻Na⁺: 0.0024 moles

in solution, you will have the mixture of a weak acid (Acetic acid), with its conjugate base (sodium acetate, CH₃COO⁻Na⁺). And pH of this buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

For Acetic buffer pKa = 4.76:

pH = 4.76 + log [CH₃COO⁻Na⁺] / [CH₃COOH]

<em>Where [] is molarity of each species or moles</em>

<em />

Replacing:

pH = 4.76 + log [0.0024 moles] / [0.0036 moles]

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3 years ago

sulfur (s) is in group 16 of the periodic table. how many valence electrons dos an atom of sulfur have?

Alex73 [517]

6 because the number of valence electrons is always the ones-place of any two-digit group number. So an element in group 18 would have 8 valence electrons
Hope this helps!

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2 years ago