2Al+6HCl⇒3H₂+2AlCl₃
<h3>Further explanation
</h3>
Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:
• 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.
• 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product
• 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
Reaction
Al+HCl⇒H₂+AlCl₃
aAl+bHCl⇒cH₂+AlCl₃
Al, left=a, right=1⇒a=1
Cl, left=b, right=3⇒b=3
H, left=b, right=2c⇒b=2c⇒3=2c⇒c=3/2
the equation becomes :
Al+3HCl⇒3/2H₂+AlCl₃ x2
2Al+6HCl⇒3H₂+2AlCl₃
CaBr2(s) Ca+2(aq)+2Br-(aq) which means, <span>Solid is turning into free ions so entropy is increasing .</span>
Answer:
<h2>The answer you are looking for is (B)</h2>
Explanation:
hope this helps
<h2>please mark as brainliest!!!</h2>
The answer is (3) An electron in the third shell has more energy than an electron in the second shell. The energy of electron will increase when number of shell increase.
<span>361.4 pm is the length of the edge of the unit cell.
First, let's calculate the average volume each atom is taking. Start with calculating how many moles of copper we have in a cubic centimeter by looking up the atomic weight.
Atomic weight copper = 63.546
Now divide the mass by the atomic weight, getting
8.94 g / 63.546 g/mol = 0.140685488 mol
And multiply by Avogadro's number to get the number of atoms:
0.140685488 * 6.022140857x10^23 = 8.472278233x10^22
Now examine the face-centered cubic unit cell to see how many atoms worth of space it consumes. There is 1 atom at each of the 8 corners and each of those atoms is shared between 8 unit cells for for a space consumption of 8/8 = 1 atom. And there are 6 faces, each with an atom in the center, each of which is shared between 2 unit cells for a space consumption of 6/2 = 3 atoms. So each unit cell consumes as much space as 4 atoms. Let's divide the number of atoms in that cubic centimeter by 4 to determine the number of unit cells in that volume.
8.472278233x10^22 / 4 = 2.118069558x10^22
Now calculate the volume each unit cell occupies.
1 cm^3 / 2.118069558x10^22 = 4.721280262x10^-23 cm^3
Let's get the cube root to get the length of an edge.
(4.721280262x10^-23 cm^3)^(1/3) = 3.61426x10^-08 cm
Now let's convert from cm to pm.
3.61426x10^-08 cm / 100 cm/m * 1x10^12 pm/m = 361.4 pm
Doing an independent search for the Crystallographic Features of Copper, I see that the Lattice Parameter for copper at at 293 K is 3.6147 x 10^-10 m which is in very close agreement with the calculated amount above. And since metals expand and contract with heat and cold, I assume the slight difference in values is due to the density figure given being determined at a temperature lower than 293 K.</span>
