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3 years ago

12

How many grams of N2 are required to react with 2.30 moles of Mg in the following process? 3 Mg + N2 → Mg3N2? (Mg = 24.3 g/mol,

N = 14.0 g/mol)

1 answer:

aivan3 [116]3 years ago

5 0

Answer: 10.73 g

Explanation:

2.3 mol Mg * (1 mol N2 / 3 mol Mg) * (14 g Mg / 1 mol g) = 10.73

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Which factors influence the salinity of ocean surface water?

In-s [12.5K]

C and D

surface water is evaporated by the sun and concentrated the ocean salt and oceans are diluted when rivers flow into them

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3 years ago

Read 2 more answers

51.7ml at 27 Celsius and 90kpa to stp

never [62]

STP is the abbreviation of standard condition for temperature and pressure which is 273.15K temperature and 1.013× 10^5 Pa pressure. Since the pressure and temperature changes, I assume the question would ask about the result of the volume. The temperature used in ideal gas should be Kelvin, so 27 Celcius would be 300.15K.
The calculation would be
PV=T
V=T/P

V2/V1= T2*P1/T1*P2
V2/V1=273.15K* 90^10^3Pa/ 300.15K * 1.013× 10^5 Pa
V2= 0.81904 * 51.7ml
V2= 42.34ml

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3 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,

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

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

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

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

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3 years ago

True or False: Jumpstart is a fast-paced method for identifying problems and solutions in a single session that can be used with

Tpy6a [65]

Answer:

<em><u>The answer is</u></em>: <u>False.</u>

<u />

Explanation:

Six Sigma is a methodology for continuous improvement that increases performance by controlling variability and focusing the process on customer specifications. It will be used within other methods such as Rummler: Brache, Scrum and TQM.

JumpStart is an educational media franchise for children, consisting mainly of educational games, produced by JumpStart Games.

<em><u>The answer is</u></em>: <u>False.</u>

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3 years ago

If a large marshmallow has a volume of 2.50 i n 3 and density of 0.242 g/c m 3 , how much would it weigh in grams? 1 i n 3 =16.3

Illusion [34]

Density of a substance is defined as the mass of the substance divided by the volume.

Density of the substance= 0.242 g cm⁻³

volume of the substance= 2.50 in³

As, 1 in³= 16.39 cm³

So, 2.50 in³= 16.39× 2.50 cm³=40.97 cm³

As ,

$Density=\frac{mass}{volume}$

Mass=volume ×Density

Mass=40.97 × 0.242

Mass=9.916 g.

4 0

3 years ago