<span>We're diluting the first solution to make the second solution. This means both solutions will contain the same amount of H2SO4 (it's the concentration of it that changes). If you look at it this way, it's easy to find the unknown mass. Call this unknown mass m.
m * 0.875 = 275 g * 0.55 --> m = 173 g
What I'm saying here is 87.5% of the mass of the first solution needs to be the same as 55% of the mass of the second solution, because you're using the same amount of H2SO4 in both.
Please mark this brainliest!
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Answer:
The correct answer is C. element
Explanation:
The sample cannot be an element because an element - or <em>elemental substance</em> - cannot be decomposed into simpler substances. Thus, it cannot be composed by differents types of atoms. For example, an element is carbon (C).
As the sample contains <u>three types of atoms</u>, it can be a compound, a molecule or a mixture, because they can be composed by different types of atoms - of different chemical elements. For example, the sample could contain the element carbon (C) combined with other elements, for example oxygen (O) or hydrogen (H), amoing others.
The atom that is oxidized : Cr
The oxidizing agent : H₃PO₄
<h3>Further explanation</h3>
Reaction
2 H₃PO₄ (aq) + 2Cr(s) → 2 CrPO₄ (aq) + 3H₂(g)
Atoms undergoing a reduction reaction (decrease in oxidation number) and an oxidation reaction (increase in oxidation number)
H⁺(in H₃PO₄) =+1
H₂=0
Cr = 0
Cr³⁺(in CrPO₄ )
the oxidizing agent.⇒which undergoes a reduction reaction and oxidizes another compound/element : H₃PO₄
Air is a mixture because it is made up of many constituents like oxygen, argon etc. it can be physically separated by fractional distillation. it doesnt have a formula unlike compounds and elements. its composition varies differently with its constituents.
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
