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3 years ago

Magnesium bromide is an ionic compound with the chemical form MgBr2 what does the 2 tell you

Chemistry

1 answer:

schepotkina [342]3 years ago

4 0

Answer:

The number of Bromine combining magnesium

Explanation:

The subscript gives the number of atom/s combining to form a compound. For this problem, 1 atom of Mg combines with 2 atoms of Br.

To form the ionic combine, Mg will lose two electrons. A bromine atom requires just one electron to have a complete octet. So, two bromine atoms will be needed to complete the bonding process.

Therefore, Mg loses two electrons to two Br atoms and the electrostatic attraction from the charges created forms the ionic bond.

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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

4 years ago

When aluminum oxidizes in air it forms aluminum oxide (al2o3) if a 51 sheet?

Ede4ka [16]

<span>Answer: mol Al2O3 x 1 mol Al/ 2 mol Al2O3= .25 mol Al The balanced equation tells us that it takes 4 moles of Al to produce 2 moles of Al2O3. 0.50 moles Al2O3 x (4 moles Al / 2 moles Al2O3) = 1.0 moles Al 1.0 moles Al x (27.0 g Al / 1 mole Al) = 27.0 g Al</span>

8 0

4 years ago

Unit conversions.(a) Convert 10,000 dynes to units of lbm ∙ ft/s^2 and lbf(b) Convert 0.2 atm to units of Kpa and lbf/in^2(c) Co

Maru [420]

Answer:

(a) 0.72 lbm· ft/s²

(b) 20.3 kPa, 2.94 lbf / in²

(d) 1.5 x 10⁻² J, 6.1 x 10⁻² cal

Explanation:

Our strategy here will be to find the conversion factors for the quantities we are asked in each part, and perform the calculations.

(a) 10,000 dynes to lbm ·ft/s²

here we are asked to convert the force of 10,000 dynes to lbm ·ft/s². Recall that F= ma ( m= mass, a = acceleration), thus

10,000 dynes = 10 g cm/s²

converting the force

10,000 g cm/s² x (1 lbm/454 g) x (1 ft / 30.48 cm ) /s² = 0.72 lbm· ft/s²

(b)

1 atm = 101.33 pa

0.2 atm x ( 101.33 kPa ) = 20.3 kPa

1 atm = 14.7 lbf / in²

0.2 atm x ( 14.7 lbf / in² /atm ) = 2.94 lbf / in²

ºF = 9/5 ºC +32

ºF = 9/5 ( 37ºC) + 32 = 98.6 ºF

K = ºC + 273

K = (37 + 273) K = 310 K

(d) 50 in²·lbm/s² to joules and calories

Since the unit in² ·lbm/s² is not that common, lets convert it using their definition.

These are energy units, and we know the energy is the force times distance. In turn force is mass times acceleration so that the units of energy are mass time distance per time squared.

Joules is the unit of energy in the metric system.

50 in² lbm/s² = 50 in²x ( 2.54 cm/in x 1m /100cm)² x (1lbm x 0.454 Kg/lbm)/s²

= 1.5 x 10⁻² Kg m²/² = 1.5 x 10⁻² J

To convert to cal it wilñl be easier to use the value in joules just calculated:

1.5 x 10⁻² J x (4.184 cal/J) = 6.1 x 10⁻² cal

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4 years ago

As the atomic numbers of the elements in column IIA increase, the size of the atoms generally

joja [24]

Answer:

Increase

Explanation:

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3 years ago

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Slav-nsk [51]

Answer:

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