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ICE Princess25 [194]
2 years ago
9

Activity In this activity, you’ll draw and count atoms to understand how nitrogen and hydrogen react to make ammonia. Begin by s

tudying this chemical equation: N2 + 3H2 → 2NH3. Then answer the following questions. Part A Question Draw the reactants using the drawing tool. Keep in mind that one molecule of nitrogen has two bonded atoms, and one molecule of hydrogen has two bonded atoms.
Chemistry
1 answer:
vredina [299]2 years ago
3 0

Answer:

Explanation:

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_Cl2 + _NaBr → _NaCl + _Br2<br><br> Answer: 1,2,2,1
nlexa [21]
1 2 2 1 is the answer so u are correct
4 0
3 years ago
What would be the minimum energy Emin required to excite a hydrogen atom from its lowest energy level
Wewaii [24]

The minimum energy required to excite a hydrogen atom from its lowest energy level is 10.2 eV.

<h3>What is excitation?</h3>

The term excitation has to do with the promotion of an electron from a lower to a higher energy level.

In this case, we are dealing with the hydrogen atom having only one electron. Thus, the minimum energy required to excite a hydrogen atom from its lowest energy level is 10.2 eV.

Learn more about energy level:brainly.com/question/17396431

#SPJ1

3 0
1 year ago
Tetrachloromethane, CCl4 is produced from the substitution reaction between methane and chlorine gas. If the rate of formation o
Korolek [52]

The rate of disappearance of chlorine gas : 0.2 mol/dm³

<h3>Further explanation</h3>

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

For reaction :

\tt aA+bB\rightarrow cC+dD

The rate reaction :

\tt -\dfrac{1}{a}\dfrac{d[-A]}{dt}= -\dfrac{1}{b}\dfrac{d[-B]}{dt}=\dfrac{1}{c}\dfrac{d[C]}{dt}=\dfrac{1}{d}\dfrac{d[D]}{dt}

Reaction for formation CCl₄ :

<em>CH₄+4Cl₂⇒CCl₄+4HCl</em>

<em />

From equation, rate of reaction = rate of formation CCl₄ = 0.05 mol/dm³

Rate of formation of  CCl₄  = reaction rate x coefficient of  CCCl₄

0.05 mol/dm³ = reaction rate x 1⇒reaction rate = 0.05 mol/dm³

The rate of disappearance of chlorine gas (Cl₂) :

Rate of disappearance of  Cl₂  = reaction rate x coefficient of  Cl₂

Rate of disappearance of  Cl₂ = 0.05 x 4 = 0.2 mol/dm³

4 0
3 years ago
A cube has the following dimensions: 7 cm x 8 cm x 10 cm and has a density of 45 g/cm3. Calculate the mass.
gizmo_the_mogwai [7]
Density = Mass / Volume
Mass = Density * Volume
Mass = 45 * 7 * 8 * 10
Mass = 25,200 grams

6 0
3 years ago
The K a of propanoic acid ( C 2 H 5 COOH ) is 1.34 × 10 − 5 . Calculate the pH of the solution and the concentrations of C 2 H 5
Zigmanuir [339]

Answer:

2.62.

Explanation:

Okay let us first write the parameters in the question in question above out. We are given the ka value of propanoic acid, C2H5COOH to be equals to 1.34 × 10^- 5. Also, we are given the value for the initial concentration of propanoic acid to be 0.441 M.

So, let us delve right into the solution to the question and we will be starting by writting the equation below;

C2H5COOH <--------> H^+ + C2H5COO^-.

Please note that this Reaction is a reversible Reaction.

Therefore, the basic things about acid is its great tendency to release Hydrogen ion in an aqeous solution.

So, we will be taken equation above and correspond it with the time and Concentration.

C2H5COOH <----> H^+ C2H5COO^-.

Initial concentration of the C2H5COOH = 0.441 M and the initial concentration of H^+ and C2H5COO^- are both zero.

So, after a time, t, concentration of C2H5COOH= 0.441 - x and at that time the concentration of H^+ and C2H5COO^- are both x and x respectively.

Hence, Ka = [C2H5COO^-] [H^+]/ C2H5COOH. -----------------------(**).

Therefore, slotting in the values from above into equation (**), we have;

1.34 × 10^-5 = [x] [x]/ [0.441 - x].

1.34 × 10^-5= x^2/ [0.441 - x].

x^2 = 1.34 × 10^-5(0.441) - 1.34 × 10^-5x.

x^2 + 1.34 × 10^-5x - 5.91× 10^-6.

x = 2.4×10^-3.

Hence, the concentration of the propanoic acid at time, t= 0.441 - 2.4 ×10^-3.

==> 0.44 M.

pH = -log [H^+].

Then, we have; pH= - log[2.4× 10^-3].

pH= 2.62.

4 0
2 years ago
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