Answer:
1.583 moles
Explanation:
Rounded Atomic Mass of Au = 197 grams
Answer:
Carbon has an atomic number of 6. That means a carbon atom has 6 protons, 6 neutrons, and 6 electrons.
Answer:
<h2>Density = 0.8 g/cm³</h2>
Explanation:
The density of an object can be found using the formula
<h3>
</h3>
From the question
mass of kerosene = 36.4 g
volume of kerosene = 45.6 mL
To find the density substitute the values into the above formula and solve
We have
<h3>
</h3>
= 0.7982
We have the final answer as
<h3>Density = 0.8 g/cm³</h3>
Hope this helps you
Answer:
A) is true
Explanation:
For the reaction:
O₂(g) + 2F₂(g) ⇄ 2OF₂(g); Kp = 2,3x10⁻¹⁵
kp is defined as:
Kp = 2,3x10⁻¹⁵ = [OF₂]²/[O₂] [F₂]²
A) If the reaction mixture initially contains only OF₂(g), then at equilibrium, the reaction mixture will consist of essentially only O₂(g) and F₂(g). <em>TRUE. </em>As the kp is 2,3x10⁻¹⁵ means per 1 of [O₂] [F₂]² you will have just 2,3x10⁻¹⁵ of [OF₂]²
B) For this equilibrium, Kc = Kp. <em>FALSE. </em>That is true just when moles of reactants are the same than moles of products. Here there are 3 moles of reactants vs 2 moles of products.
C) If the reaction mixture initially contains only OF₂(g), then the total pressure at equilibrium will be less than the total initial pressure. <em>FALSE. </em>Because per 2 moles of OF₂(g) you will produce 3 moles of gas increasing pressure.
D) If the reaction mixture initially contains only O₂(g) and F₂(g), then at equilibrium, the reaction mixture will consist of essentially only OF₂(g). <em>FALSE. </em>For the same reason of A), the mixture will contains essentially only O₂(g) and F₂(g)
E) If the reaction mixture initially contains only O₂(g) and F₂(g), then the total pressure at equilibrium will be greater than the total initial pressure. <em>FALSE. </em>If mixture initially contains only O₂(g) and F₂(g), 3 moles will of gas will react to produce 2 moles of gas doing pressure decreases.
I hope it helps!
Answer:
a. Ca(s) + N₂(g) + 3O₂(g) → Ca(NO₃)₂(s).
b. C(s,graphite) + 2H₂(g) + 1/2O₂(g) → CH₃OH(l).
c. 3NaCl(s) + 4O₃(g) → 3NaClO₄(s).
Explanation:
<em>a. The formation reaction of Ca(NO₃)₂(s) is:</em>
Ca(s) + N₂(g) + 3O₂(g) → Ca(NO₃)₂(s).
- 1.0 mole of solid Ca reacts with 1.0 mole of N₂ gas and 3.0 moles of O₂ gas to produce 1.0 mole of solid Ca(NO₃)₂.
<em>b. The formation reaction of CH₃OH(l) is:</em>
C(s,graphite) + 2H₂(g) + 1/2O₂(g) → CH₃OH(l).
- 1.0 mole of solid C (graphite) reacts with 2.0 mole of H₂ gas and 0.5 mole of O₂ gas to produce 1.0 mole of liquid CH₃OH.
<em>c. The formation reaction of NaClO₄(s) is:</em>
3NaCl(s) + 4O₃(g) → 3NaClO₄(s).
- 3.0 moles of solid NaCl react with 4.0 moles of O₂ gas to produce 3.0 moles of solid NaClO₄.