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2 years ago

which are correct formulas for ionic compounds? for those that are not correct, give the correct formula and justify your answer

Chemistry

1 answer:

galina1969 [7]2 years ago

4 0

Answer:

a. AlCl is incorrect because Al has a +3 charge while Cl only has a -1 charge. The correct formula would be AlCl₃. This balances the charges.

b. Na₃SO₄ is incorrect because Na has a charge of +1 and there are three of them so its +3 and SO₄ has a charge of -2. The correct formula would be Na₂SO₄. This balances the charges.

c. BaOH₂ is incorrect because the polyatomic ion OH would not be written that way. It would be written like this Ba(OH)₂. Writing it like BaOH₂ gives the impression that instead of having 2 OH it has 2 H and 1 O.

d. Fe₂O is incorrect because Fe either has a +2 charge or +3 charge while O has a -2 charge. Possible correct answers could be FeO (iron (II) oxide) or Fe₂O₃ (iron (III) oxide).

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How many atoms are in 57.5 liters of xenon gas at STP

TiliK225 [7]

Answer:

1.55 × 10²⁴ atoms Xe

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Moles
STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 57.5 L Xe at STP

[Solve] atoms Xe

Step 2: Identify Conversions

[STP] 22.4 L = 1 mol

Avogadro's Number

Step 3: Convert

[DA] Set up: $\displaystyle 57.5 \ L \ Xe(\frac{1 \ mol \ Xe}{22.4 \ L \ Xe})(\frac{6.022 \cdot 10^{23} \ atoms \ Xe}{1 \ mol \ Xe})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 1.54583 \cdot 10^{24} \ atoms \ Xe$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1.54583 × 10²⁴ atoms Xe ≈ 1.55 × 10²⁴ atoms Xe

4 0

3 years ago

When you slide a box across the floor what force must your push be stronger than?

weeeeeb [17]

If I remember correctly, the answer is Friction force.

6 0

3 years ago

What is the mass of 3.8 × 10^24 atoms of argon (ar)? 0.0039 g 0.16 g 6.3 g 250 g

ale4655 [162]

Atomic mass Ar => 39.948 a.m.u

39.948 g --------------- 6.02x10²³ atoms
?? g -------------------- 3.8x10²⁴ atoms

(3.8x10²⁴) x 39.948 / 6.02x10²³ => 250 g

hope this helps!

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3 years ago

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FrozenT [24]

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3 years ago

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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