1. B
The positive charge in water is provided by hydrogen, and gold provides the same charge. However, gold is not more reactive than hydrogen so it can not replace it in the compound.
2. In order to balance the equation, you must sure there are equal moles of each element on the left and right side of the equation:
2C₂H₆ + 7O₂ → 4CO₂ + ₆H₂O
3. The number of moles of sodium atoms on the left of the equation must be equal to the number of moles of sodium atoms on the right, as per the law of conservation of mass. The answer is B.
4. C.
A synthesis reaction usually results from single displacement because some element or compound is produced in its pure form
5. B.
The gas being produced is being synthesized.
Answer:
Explanation:
We want to convert from moles to grams, so we must use the molar mass.
<h3>1. Molar Mass</h3>
The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).
We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.
- Phosphorus (P): 30.973762 g/mol
- Iodine (I): 126.9045 g/mol
Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.
- I₃: 126.9045 * 3=380.7135 g/mol
- PI₃: 30.973762 + 380.7135 = 411.687262 g/mol
<h3>2. Convert Moles to Grams</h3>
Use the molar mass as a ratio.
We want to convert 3.14 moles to grams, so we multiply by that value.
The units of moles of PI₃ cancel.
<h3>3. Round</h3>
The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.
The 2 in the ones place tells us to leave the 9.
3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>
<u>Answer:</u>
<u>For a:</u> The isotopic representation of iodine is
<u>For b:</u> The isotopic representation of cesium is
<u>For c:</u> The isotopic representation of strontium is
<u>Explanation:</u>
The isotopic representation of an atom is:
where,
Z = Atomic number of the atom
A = Mass number of the atom
X = Symbol of the atom
We are given:
Number of neutrons = 78
Atomic number of iodine = 53 = Number of protons
Mass number = 53 + 78 = 131
Thus, the isotopic representation of iodine is
We are given:
Number of neutrons = 82
Atomic number of cesium = 55 = Number of protons
Mass number = 55 + 82 = 137
Thus, the isotopic representation of cesium is
We are given:
Number of neutrons = 52
Atomic number of strontium = 38 = Number of protons
Mass number = 38 + 52 = 90
Thus, the isotopic representation of strontium is