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3 years ago

A student is building an electrical circuit. Which material should she choose for wires, and why.

Chemistry

1 answer:

r-ruslan [8.4K]3 years ago

3 0

D. A metal such as copper, because its atom have very mobile electrons - A-p-e-x verified!

<u>Have a Great Day!</u>

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The concentration of a saturated BaCl2 solution is 1.75 M (mol/liter) and the concentration of a saturated Na2SO4 solution is 2.

Kitty [74]

Answer:

a) The theoretical yield is 408.45g of $BaSO_{4}$

b) Percent yield = $\frac{realyield}{408.45g}*100$

Explanation:

1. First determine the numer of moles of $BaCl_{2}$ and $Na_{2}SO_{4}$ .

Molarity is expressed as:

M= $\frac{molessolute}{Lsolution}$

- For the $BaCl_{2}$

M= $\frac{1.75molesBaCl_{2}}{1Lsolution}$

Therefore there are 1.75 moles of $BaCl_{2}$

- For the $Na_{2}SO_{4}$

M= $\frac{2.0moles[tex]Na_{2}SO_{4}$ }{1Lsolution}[/tex]

Therefore there are 2.0 moles of $Na_{2}SO_{4}$

2. Write the balanced chemical equation for the synthesis of the barium white pigment, $BaSO_{4}$ :

$BaCl_{2}+Na_{2}SO_{4}=BaSO_{4}+2NaCl$

3. Determine the limiting reagent.

To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:

- For the $BaCl_{2}$ :

$\frac{1.75}{1}=1.75$

- For the $Na_{2}SO_{4}$ :

$\frac{2.0}{1}=2.0$

As the $BaCl_{2}$ is the smalles quantity, this is the limiting reagent.

4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

$1.75molesBaCl_{2}*\frac{1molBaSO_{4}}{1molBaCl_{2}}*\frac{233.4gBaSO_{4}}{1molBaSO_{4}}=408.45gBaSO_{4}$

5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:

Percent yield = $\frac{realyield}{theoreticalyield}*100$

Percent yield = $\frac{realyield}{408.45g}*100$

The real yield is the quantity of barium white pigment you obtained in the laboratory.

7 0

3 years ago

What is 5 divided by 2.7

Strike441 [17]

5 divided by 2.7 is 0.54

3 0

2 years ago

Which value of x from the set {4, 5, 6, 7), makes this equation true?

irina [24]

C and d is the answer

3 0

3 years ago

Read 2 more answers

12a. How does the model of the volcano provide evidence for the cycling of Earth materials through the rock cycle?

Bond [772]

Answer:

lava is coming from the crust

Explanation:

volcano's explode with lava because the crust pushes it out

6 0

2 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago