1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
luda_lava [24]
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)
Chemistry
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

You might be interested in
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
4 0
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
6 0
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}.

8 0
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>

3 0
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
Other questions:
  • The equilibrium constant for A + 2B → 3C is 2.1 * 10^-6
    15·1 answer
  • Calculate the volume of this irregular solid to the nearest cubic centimeter​
    7·2 answers
  • This is no help<br> Ok so bydhcvx
    13·2 answers
  • At a pressure of 782.3 mm Hg and 34.4 °C, a certain gas has a volume of 362.4 mL. What will
    5·1 answer
  • Please help me :')<br>1. D<br>2. A and D<br>3. A<br>4. C and D<br>5. B<br>6. A and B<br>7. C​
    12·1 answer
  • If you were going to be a scientist, what type of scientist would you be?
    11·2 answers
  • How many mol of hydrogen are needed to double the volume occupied by 0.34 mol of hydrogen,
    8·1 answer
  • A liquid's freezing point is -38°C and it's boiling point is 357°C. What is the number of Kelvin between the boiling point and t
    6·1 answer
  • Whats the molar mass of c6h6
    13·1 answer
  • Use the periodic table and example bond table to answer the question.
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!