Answer:
c
Explanation:
How many moles of gold are equivalent to 1.204 × 1024 atoms?
0.2
0.5
2
5
C) 2 Is the correct answer, I took the test and it was correct.
Taking into account the reaction stoichiometry, you can observe that:
- one mole of Ca₃P₂ produces 2 mol of PH₃.
- the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
Ca₃P₂ + 6 H₂O → 3 Ca(OH)₂ + 2 PH₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Ca₃P₂:1 mole
- H₂O: 6 moles
- Ca(OH)₂: 3 moles
- PH₃: 2 moles
The molar mass of the compounds is:
- Ca₃P₂: 182 g/mole
- H₂O: 18 g/mole
- Ca(OH)₂: 74 g/mole
- PH₃: 34 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Ca₃P₂: 1 mole ×182 g/mole= 182 grams
- H₂O: 6 moles× 18 g/mole= 108 grams
- Ca(OH)₂: 3 moles ×74 g/mole= 222 grams
- PH₃: 2 moles ×34 g/mole= 68 grams
<h3>Correct statements</h3>
Then, by reaction stoichiometry, you can observe that:
- one mole of Ca₃P₂ produces 2 mol of PH₃.
- the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.
Learn more about the reaction stoichiometry:
<u>brainly.com/question/24741074</u>
<u>brainly.com/question/24653699</u>
Its most significant when all other forces are absent
Carbohydrates. If you think about it, it's a mix of the three words.
According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate).
(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt
(10.0 g) / (0.0480 mol) = 208.3 g/mol
So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71
