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VLD [36.1K]
3 years ago
12

What are the amount of moles in 12.15 grams of Magnesium?

Chemistry
1 answer:
zaharov [31]3 years ago
8 0
I think the amount would be a 0.4998 mol
I did moles=mass(g)/A,r
=12.5/24.3 to get that
You might be interested in
Which possible discovery would be attributed to an inorganic molecule?
egoroff_w [7]

Answer:

fireproofing spray used in firefighting can be termed as one of the possible discovery that would be attributed to an inorganic molecule.

Explanation: • There a number of advancement been made each day to improve the different process either domestic, industrial, or related to research work. As a number of professionals are always in search to get the right set of information or data to get to a more proper conclusion.

4 0
2 years ago
Read 2 more answers
A PNP transistor is connected in a circuit so that the collector-base junction remains reverse biased and the emitter-base junct
maw [93]
Here are the choices:

A. the output current will be much smaller than the input current. 
<span>B. the output voltage will be much larger than the input voltage. </span>
<span>C. the output voltage will be much smaller than the input voltage. </span>
<span>D. the output current will be much larger than the input current
</span>
The correct answer is letter <span>B. the output voltage will be much larger than the input voltage.</span>
5 0
4 years ago
A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch
ivanzaharov [21]

Answer:

The reactions free energy \Delta G = -49.36 kJ

Explanation:

From the question we are told that

      The pressure of (NO) is P_{NO} = 9.20 \ atm

      The  pressure of  (Cl) gas is  P_{Cl} = 9.15 \ atm

       The  pressure of nitrosly chloride (NOCl) is P_{(NOCl)} = 7.70 \ atm

The reaction is

              2NO_{(g)} + Cl_2 (g)    ⇆   2 NOCl_{(g)}

 From the reaction we can  mathematically evaluate the \Delta G^o (Standard state  free energy ) as

                    \Delta G^o = 2 \Delta G^o _{NOCl} -   \Delta G^o _{Cl_2}  - 2 \Delta G^o _{NO}

The Standard state  free energy for NO is  constant with a value  

                 \Delta G^o _{NO} = 86.55 kJ/mol

 The Standard state  free energy for Cl_2 is  constant with a value                  

             \Delta G^o _{Cl_2} = 0kJ/mol

 The Standard state  free energy for NOCl is  constant with a value

         \Delta G^o _{NOCl} =66.1kJ/mol

Now substituting this into the equation

        \Delta G^o = 2 * 66.1 - 0 - 2 * 87.6

                = -43 kJ/mol

The pressure constant is evaluated as

         Q =  \frac{Pressure \ of  \ product }{ Pressure  \ of \ reactant }

Substituting  values  

        Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}

           = 0.0765

The free energy for this reaction is evaluated as

           \Delta  G  =  \Delta  G^o  + RT ln Q

Where R is gas constant with a value  of  R = 8.314 J / K \cdot mol

          T is temperature in K  with a given value of  T = 25+273 = 298 K

   Substituting value

                \Delta  G  = -43 *10^{3} + 8.314 *298 * ln [0.0765]

                       = -43-6.36

                      \Delta G = -49.36 kJ

4 0
3 years ago
Determine the pH of the resulting solution if 25 mL of 0.400 M strychnine (C21H22N2O2) is added to 50 mL of 0.200 M HCl? Assume
DIA [1.3K]

Answer:

pH = 4.56

Explanation:

The strychnine reacts with HCl as follows:

C₂₁H₂₂N₂O₂ + HCl ⇄ C₂₁H₂₂N₂O₂H⁺ + Cl⁻

<em />

For strychnine buffer:

pOH = 5.74 + log [C₂₁H₂₂N₂O₂H⁺] / [C₂₁H₂₂N₂O₂]

Initial moles of C₂₁H₂₂N₂O₂ are:

0.025L * (0.400 mol / L) = 0.01 moles C₂₁H₂₂N₂O₂

And of HCl are:

0.05L * (0.200 mol / L) = 0.01 moles HCl

That means after the reaction, you will have just 0.01 moles of C₂₁H₂₂N₂O₂H⁺ in 50mL + 25mL = 0.075L. And molarity is:

[C₂₁H₂₂N₂O₂H⁺] = 0.01 mol / 0.075L = 0.1333M

This conjugate acid, is in equilibrium with water as follows:

C₂₁H₂₂N₂O₂H⁺(aq) + H₂O(l) ⇄ C₂₁H₂₂N₂O₂ + H₃O⁺

<em />

<em>Where Ka = Kw / Kb = 1x10⁻¹⁴ / 1.8x10⁻⁶ = 5.556x10⁻⁹</em>

<em />

Ka is defined as:

Ka = 5.556x10⁻⁹ = [C₂₁H₂₂N₂O₂] [H₃O⁺] / [C₂₁H₂₂N₂O₂H⁺]

In equilibrium, concentrations are:

C₂₁H₂₂N₂O₂ = X

H₃O⁺ = X

C₂₁H₂₂N₂O₂H⁺ = 0.1333M - X

Replacing in Ka expression:

5.556x10⁻⁹ = [X] [X] / [0.1333M - X]

7.39x10⁻¹⁰ - 5.556x10⁻⁹X = X²

7.39x10⁻¹⁰ - 5.556x10⁻⁹X - X² = 0

Solving for X:

X = - 2.72x10⁻⁵M → False solution. There is no negative concentrations

X = 2.72x10⁻⁵M → Right solution.

As H₃O⁺ = X

H₃O⁺ = 2.72x10⁻⁵M

And pH = -log H₃O⁺

<h3>pH = 4.56</h3>
4 0
3 years ago
Why do warm currents begin near the equator?
r-ruslan [8.4K]

Answer:

I would go with A

Explanation:

Because the earths equator is warmed by most direct rays of the sun, air a the equator is hotter than air further north or the south. The hotter air rises up at the equator and as colder air moves in to take its place, the wind begins to blow and push the ocean into waves and currents

5 0
3 years ago
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