The reaction produces 2.93 g H₂.
M_r: 133.34 2.016
2Al + 6HCl → 2AlCl₃ + 3H₂
<em>Moles of AlCl₃</em> = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃
<em>Moles of H₂</em> = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂
<em>Mass of H₂</em> = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂
Answer:
A
Explanation:
Hmm, so we have the following in the diagram
Pt(s)
Cl2(g)
Ag(s)
NaCl(aq)
AgNO3(aq)
Pt 2+, 4+, 6+ Though it states Pt is inert
Cl 2-
Ag 1+
Na 1+
NO3-
Anode definition: the positively charged electrode by which the electrons leave an electrical device.
Electrode definition: a conductor through which electricity enters or leaves an object, substance, or region.
Cations attracted to cathode pick up electrons
Anions attracted to anode release electrodes+
Reduction at Cathode (red cat gain of e)
Oxidation at Anode (ox anode loss of e)
So from the diagram we can see that the charge is being generated through the 2 metal plates.
So the answer is A, the anode material is Pt and the half reaction is 2Cl- = Cl2 + 2e-
Answer: 0.151
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
![Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1d%5BHBr%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7B1d%5BH_2%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D)
Given: ![-\frac{d[HBr]}{dt}]=0.301](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D%5D%3D0.301)
Putting in the values we get:
![\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B0.301%7D%7B2%7D%3D%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D)
![+\frac{1d[Br_2]}{dt}=0.151](https://tex.z-dn.net/?f=%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D%3D0.151)
Thus the rate of appearance of 
is 0.151
<h3>
Answer:</h3>
56.11 g/mol
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Compound] KOH
<u>Step 2: Identify</u>
[PT] Molar Mass of K - 39.10 g/mol
[PT] Molar Mass of O - 16.00 g/mol
[PT] Molar Mass of H - 1.01 g/mol
<u>Step 3: Find</u>
39.10 + 16.00 + 1.01 = 56.11 g/mol
<u>Answer:</u>
211.9 J
<u>Explanation:</u>
The molecules of water release heat during the transition of water vapor to liquid water, but the temperature of the water does not change with it.
The amount of heat released can be represented by the formula:
where 
= heat energy, 
= mass of water and 
= latent heat of evaporation.
The latent heat of evaporation for water is 
and the mass of the water is 
.
The amount of heat released in this process is:
211.9 J
