Molarity of 275 mL of solution containing 135 mol of glucose.
<h3>What is molarity?</h3>
Molar concentration (also known as molarity, quantity concentration, or substance concentration) is a measure of the concentration of a chemical species in a solution, specifically of a solute, in terms of amount of substance per unit volume of solution. The most often used unit for molarity in chemistry is the number of moles per liter, denoted by the unit symbol mol/L or mol/dm3 in SI units. A solution with a concentration of 1 mol/L is referred to as 1 molar, or 1 M.
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Answer:
a. 9.2
b. 4.4
c. 6.3
Explanation:
In order to calculate the pH of each solution, we will use the definition of pH.
pH = -log [H⁺]
(a) [H⁺] = 5.4 × 10⁻¹⁰ M
pH = -log [H⁺] = -log 5.4 × 10⁻¹⁰ = 9.2
Since pH > 7, the solution is basic.
(b) [H⁺] = 4.3 × 10⁻⁵ M
pH = -log [H⁺] = -log 4.3 × 10⁻⁵ = 4.4
Since pH < 7, the solution is acid.
(c) [H⁺] = 5.4 × 10⁻⁷ M
pH = -log [H⁺] = -log 5.4 × 10⁻⁷ = 6.3
Since pH < 7, the solution is acid.
Answer:
The nitrile group
Explanation:
The nitrile group contains the C≡N bond. It should be recalled that triple bond is highly electronegative and withdraws electrons from the C-H bond more effectively than the halogen atom.
The higher effectiveness of the C≡N bond at electron withdrawal greatly reduces the electron density of the C-H bond thereby making the hydrogen atom of the bond highly labile
I think that a writer may use all of the following techniques to lead the reader to a flashback EXCEPT: a. a memory
Answer:
B) 2Crº + 6e- --> 2Cr3+
Explanation:
The process of oxidation is where electrons are lost. Thus, out of the 2 ions that change charge(Cr and Cu), we must choose the one where the oxidation number increases(which means electrons are lost). Cr goes from an oxidation number of 0 to an oxidation number of 3+, while Cu goes from an oxidation number of 2+ to 0. Thus, we are looking at the half reaction for Cr. Half reactions never have subtracting electrons, so the answer must be B. I am assuming that last plus should be a -->