Answer:
0.57 moles (NH4)3PO4 (2 sig. figs.)
Explanation:
To quote, J.R.
"Note: liquid ammonia (NH3) is actually aqueous ammonium hydroxide (NH4OH) because NH3 + H2O -> NH4OH.
H3PO4(aq) + 3NH4OH(aq) ==> (NH4)3PO4 + 3H2O
Assuming that H3PO4 is not limiting, i.e. it is present in excess
1.7 mol NH4OH x 1 mole (NH4)3PO4/3 moles NH4OH = 0.567 moles = 0.57 moles (NH4)3PO4 (2 sig. figs.)"
Answer:
anciet fossils
Explanation:
Radiocarbon dating involves determining the age of an ancient fossil or specimen by measuring its carbon-14 content. Carbon-14, or radiocarbon, is a naturally occurring radioactive isotope that forms when cosmic rays in the upper atmosphere strike nitrogen molecules, which then oxidize to become carbon dioxide.
Cylinder A - Adding 25 grams raised the water level by roughly 3.2 mL
Cylinder B - Adding 25 grams raised the water level by roughly 2.8 mL
Cylinder C - Adding 25 grams raised the water level by roughly 3.5 mL
Calculate grams/mL (density)
A. 25 g/3.2 mL = 7.8125 g/mL = Iron
B. 25 g/2.8 mL = 8.9286 g/mL = Nickel
C. 25 g/3.5 mL = 7.1429 g/mL = Zinc
Hope I helped!
Answer:
The answer to your question is [H₃O⁺] = 0.025 [OH⁻] = 3.98 x 10⁻¹³
Explanation:
Data
[H⁺] = ?
[OH⁻] = ?
pH = 1.6
Process
Use the pH formula to calculate the [H₃O⁺], then calculate the pOH and with this value, calculate the [OH⁻].
pH formula
pH = -log[H₃O⁺]
-Substitution
1.6 = -log[H₃O⁺]
-Simplification
[H₃O⁺] = antilog (-1,6)
-Result
[H₃O⁺] = 0.025
-Calculate the pOH
pOH = 14 - pH
-Substitution
pOH = 14 - 1.6
-Result
pOH = 12.4
-Calculate the [OH⁻]
12.4 = -log[OH⁻]
-Simplification
[OH⁻] = antilog(-12.4)
-Result
[OH⁻] = 3.98 x 10⁻¹³
Answer:
C IS YOUR CORRECT ANSWER
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