Answer:
Explanation:
CH₃COOH + NaOH = CH₃COONa + H₂O .
42.5 mL of .115 M of NaOH will contain .0425 x .115 moles of NaOH
= 48.875 x 10⁻⁴ moles NaOH
It will react with same number of moles of acetic acid
So number of moles of acetic acid in 3.45 mL = 48.875 x 10⁻⁴
number of moles of acetic acid in 1000 mL = 48.875 x 10⁻⁴ x 10³ / 3.45 moles
= 1.4167 moles
= 1.4167 x 60 gram
= 85 grams .
So 85 grams of acetic acid will be contained in one litre of acetic acid.
In NaMnO₄, Mn has the highest oxidation number.
The question is incomplete, the complete question is;
Which of the following species contains manganese with the highest oxidation number?
A) Mn
B) MnF₂
C) Mn₃(PO₄)₂
D) MnCl₄
E) NaMnO₄
In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.
1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.
2) For MnF₂;
Mn has an oxidation number of +2
3) For Mn₃(PO₄)₂
Mn has an oxidation number of +2
4) For MnCl₄
Mn has an oxidation number of +4
5) For NaMnO₄
Mn has an oxidation number of +7
Hence in NaMnO₄, Mn has the highest oxidation number.
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Answer:
26.0 g/mol is the molar mass of the gas
Explanation:
We have to combine density data with the Ideal Gases Law equation to solve this:
P . V = n . R .T
Let's convert the pressure mmHg to atm by a rule of three:
760 mmHg ____ 1 atm
752 mmHg ____ (752 . 1)/760 = 0.989 atm
In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.
Moles = Mass / molar mass.
We can replace density data as this in the equation:
0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K
(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x
0.0405 mol = 1.053 g / x
x = 1.053 g / 0.0405 mol = 26 g/mol
The appropriate response is oxygen. Cellular respiration is the procedure cells use to make vitality. Cells in our body join glucose and oxygen to make ATP and carbon dioxide. Oxygen is utilized as an electron acceptor inside the electron transport tie of vigorous breath to create adenosine triphosphate or ATP. This compound is a fundamental part in intracellular vitality exchange.
Answer:
The relationship is the "<em>variation</em><em> </em><em>of</em><em> </em><em>vapour</em><em> </em><em>pressure</em><em> </em><em>with</em><em> </em><em>temperature</em><em>"</em><em> </em>or <em>"</em><em> </em><em>the</em><em> </em><em>effect</em><em> </em><em>of</em><em> </em><em>vapour</em><em> </em><em>pressure</em><em> </em><em>on</em><em> </em><em>the</em><em> </em><em>melting</em><em> </em><em>and</em><em> </em><em>boiling</em><em> </em><em>points</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>phase</em><em>"</em>