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Explanation:
Answer:
0.500 mole of Xe (g) occupies 11.2 L at STP.
General Formulas and Concepts:
<u>Gas Laws</u>
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Stoichiometry</u>
- Mole ratio
- Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
<em>Identify.</em>
0.500 mole Xe (g)
<u>Step 2: Convert</u>
- [DA] Set up:

- [DA] Evaluate:

Topic: AP Chemistry
Unit: Stoichiometry
Volume of a substance can be determined by dividing mass of the substance by its density.
That can be mathematical shown as:
Density=Mass/Volume
So, Volume=Mass/Density
Here mass of the substance given as 24.60 g
Whereas density of the substance is 2.70 g/mL
So,
Volume=Mass/Density
=24.6/2.7
=9.1 mL
So volume of the substance is 9.1 mL.
Answer:
a) AgNO3 + KI → AgI + KNO3
b) Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O
c) 2Na3PO4 + 3Ni(NO3)2 → Ni3(PO4)2 + 6NaNO3
d) 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Explanation:
a) AgNO3 + KI → Ag+ + NO3- + K+ + I-
Ag+ + NO3- + K+ + I- → AgI + KNO3
AgNO3 + KI → AgI + KNO3
b) Ba(OH)2 + 2HNO3 → Ba^2+ + 2OH- + 2H+ + 2NO3-
Ba^2+ + 2OH- + 2H+ + 2NO3- → Ba(NO3)2 + 2H2O
Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O
c) 2Na3PO4 + 3Ni(NO3)2 → 6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3-
6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3- → Ni3(PO4)2 + 6NaNO3
2Na3PO4 + 3Ni(NO3)2 → Ni3(PO4)2 + 6NaNO3
d) 2Al(OH)3 + 3H2SO4 → 2Al^3+ + 6OH- + 6H+ + 3SO4^2-
2Al^3+ + 3OH- + 3H+ + 3SO4^2- → Al2(SO4)3 + 6H2O
2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O