Answer:
The alteration in the state of matter occurs when sufficient thermal energy is either added to or taken away from an object.
Answer:
Explanation:
AgNO3 + NaCl --> AgCl + NaNO3
Moles
of AgNO3
= molarity * volume
= 1 * 0.01
= 0.01 mol
for NaCl
= 0.01 * 1
= 0.01 mol.
According to stoichiometry, one mole of silver nitrate corresponds to one mole of NaCl reacted. Hence,
Moles of AgCl generated = 0.01 × 1
= 0.01 mol AgCl produced.
Heat gained by the solution as precipitation occurs:
Solution mass = density × volume
= 1 × 20
= 20 g.
Using q = m * Cp * (T2 - T1)
= 20 * 4.18 * (32.6 - 25.0)
= 635 J
The absorbed heat of 635 J indicates the reaction released -635 J
Thus, Delta H = -635 J/0.01 mol
= -63500 J/mol
= -63.5 kJ/mol.
Answer:
0.0181 mol H
Explanation:
Step 1: Provided data
Moles of water (H₂O) produced during the process: 0.00905 mol H₂O
Step 2: Determine the quantity of H (in mol) present in 0.00905 mol of H₂O
According to the H2O chemical formula, the ratio of water to hydrogen is 1:2, meaning 2 moles of H exist for every mole of H₂O. This conversion factor will be applied to find the moles of H contained in 0.00905 moles of H₂O.
0.00905 mol H₂O × (2 mol H/1 mol H₂O) = 0.0181 mol H
1.09 moles of tin (Sn)
Explanation:
The reaction we are examining involves tin chloride (SnCl₂) reacting with sodium (Na) to yield tin (Sn) and sodium chloride (NaCl):
SnCl₄ + 4 Na → Sn + 4 NaCl
Considering the chemical equation, this results in:
If 1 mole of SnCl₂ reacts with 4 moles of Na,
then 2.34 moles of SnCl₂ will react with X moles of Na.
X = (2.34 × 4) / 1 = 9.36 moles of Na
So, 2.34 moles of SnCl₂ can react with 9.36 moles of Na, but we have only 4.36 moles of Na on hand. Therefore, Na is the limiting reactant. Based on this, we can conclude:
If 4 moles of Na yield 1 mole of Sn,
then 4.36 moles of Na will produce Y moles of Sn.
Y = (4.36 × 1) / 4 = 1.09 moles of Sn
Learn more about:
limiting reactant
To calculate the average atomic mass of an element, you can multiply the masses of its isotopes by their corresponding relative abundances and sum them up.
Average atomic mass of Br = 158 amu(0.2569) + 160 amu(0.4999) + 162 amu(0.2431)
Average atomic mass = 159.96 amu
According to the information given, the relative abundance of Br-79 stands at 25.69% because two atoms of Br yield 79*2 = 158 amu. Likewise, for Br-81, the relative abundance is 162 due to 81*2 = 162, which is 24.31%.
