The equation that shows the formation of chromium (ii) ion from neutral chromium atom is as follow
Cr ---> cr^2+ + 2e-
Cr^2+ is the chromium ion with oxidation state of two which is one of the common ion of chromium. Other common ion of chromium include chromium of oxidation state 6 and 3
Answer:
a) distance is 4+7+1+8=20 blocks
b) displacement is 10 blocks
Explanation:
find displacement: x and y
x axis displacement = 4-1 = 3 blocks
y axis displacement = -7+8= 1 block
displacement = the square root of 3^2 + 1^2
= 9+1 = 10 blocks.
You can find the angle of displacement with respect to the initial position using trig identities, if you wish.
Answer:
The answer to your question is 8.21 g of H₂O
Explanation:
Data
mas of water = ?
mass of hydrogen = 4.6 g
mass of oxygen = 7.3 g
Balanced chemical reaction
2H₂ + O₂ ⇒ 2H₂O
Process
1.- Calculate the atomic mass of the reactants
Hydrogen = 4 x 1 = 4 g
Oxygen = 16 x 2 = 32 g
2.- Calculate the limiting reactant
Theoretical yield = H₂/O₂ = 4 / 32 = 0.125
Experimental yield = H₂/ O₂ = 4.6/7.3 = 0.630
From the results, we conclude that the limiting reactant is Oxygen because the experimental yield was higher than the theoretical yield.
3.- Calculate the mass of water
32 g of O₂ ---------------- 36 g of water
7.3 g of O₂ --------------- x
x = (7.3 x 36) / 32
x = 262.8 / 32
x = 8.21 g of H₂O
Answer: 17.34 grams of alum will be produced if 0.9875 g of Aluminium foil was used.
Explanation: Reaction to form alum from Aluminium is given as:

We are given Aluminium to be the limiting reactant, so the formation of alum will be dependent on Aluminium because it limits the formation of product.
By stoichiometry,
2 moles of Al is producing 2 moles of Alum
Mass of 2 moles of Aluminium = (2 × 27)g/mol = 54 g/mol
Mass of 2 moles of alum = (2 × 474)g/mol = 948 g/mol
54 g/mol of aluminium will produce 948 g/mol of alum, so

Amount of Alum produced = 17.34 grams
Theoretical yield of alum = 17.34 grams.
Answer:
0.264 mol/L
Explanation:
Data obtained from the question includes:
Volume = 5.74L
Mass of FeCl3 = 246 g
Molarity =?
Next, we shall determine the number of mole in 246g of FeCl3.
This is illustrated below:
Mass of FeCl3 = 246g
Molar mass of FeCl3 = 56 + (35.5x3) = 162.5g/mol
Number of mole of FeCl3 =..?
Mole = Mass /Molar Mass
Number of mole of FeCl3 = 246/162.5
Number of mole of FeCl3 = 1.514 mol
Now, we can obtain the molarity of the solution as follow:
Molarity = mole /Volume
Volume = 5.74L
Mole of FeCl3 = 1.514 mol
Molarity = 1.514/5.74
Molarity = 0.264 mol/L