Answer:
oxide of potassium or potassium oxide i believe.
Answer:
The pH of the buffer solution = 8.05
Explanation:
Using the Henderson - Hasselbalch equation;
pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]
where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21
Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)
[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M
[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M
Therefore,
pH = 7.21 + log (0.663 / 0.096)
pH = 7.21 + 0.84
pH = 8.05
A. Combustion since it can’t be nuclear….?
The information that a foliated metamorphic rock provide about the conditions it was formed under are:(for many diffrent rocks like Phyllite and Amphibolite)
you can tell by its features like if its smooth and sleek or if its rough and bumpy those physical features.
Answer is: 1.29 grams <span>of solid formed.
</span>Chemical reaction: 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq).
n(AgNO₃) = c(AgNO₃) · V(AgNO₃).
n(AgNO₃) = 0.220 M · 0.0351 L.
n(AgNO₃) = 0.0078 mol; limiting reactant.
n(K₂CrO₄) = 0.420 M · 0.052 L.
n(K₂CrO₄) = 0.022 mol.
From chemical reaction: n(AgNO₃) : n(Ag₂CrO₄) = 2 : 1.
n(Ag₂CrO₄) = 0.0078 mol ÷ 2.
n(Ag₂CrO₄) = 0.0039 mol.
m(Ag₂CrO₄) = 0.0039 mol · 331.73 g/mol.
m(Ag₂CrO₄) = 1.29 g.
