A 100 milliliters (mL) sample of a liquid is poured from a beaker into a graduated cylinder. Which of the following best explain

HNO₃ → H⁺ + NO₃⁻

v₁=35.0 mL
c₁=0.255 mmol/mL
n₁(NO₃⁻)=v₁c₁

Mg(NO₃)₂ → Mg²⁺ + 2NO₃⁻

v₂=45.0 mL
c₂=0.328 mmol/mL
n₂(NO₃⁻)=2c₂v₂

c₃={n₁+n₂}/(v₁+v₂)={c₁v₁ + 2c₂v₂}/(v₁+v₂)

c₃={35.0*0.255+2*0.328*45.0}/(35.0+45.0)≈0.481 mmol/mL

a. 0.481 m

Answer:

The solution to your inquiry is C = 0.000333 kcal/g°C

or C = 0.333 cal/g°C

Explanation:

Data

Q = 1.67 kcal

mass = 79.2 g

ΔT = 63.3°C

Formula

Q = mCΔT

Solving for C

C = Q/mΔT

Substituting values

C = 1.67/(79.2 x 63.3)

Simplifying

C = 1.67 / 5013.4

Final Result

C = 0.000333 kcal/g°C

or C = 0.333 cal/g°C

Every unicellular organism prospers by executing metabolic activities.

Metabolic activities encompass the set of chemical reactions essential for sustaining life.

Explanation:

Different metabolic pathways maintain an organism's viability. Various metabolic activities occur in all living organisms.

These include processes like cellular respiration, reproduction, excretion, and digestion. Each living cell engages in these activities to survive.

Organisms acquire the energy necessary for these activities through food consumption.

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Response:

The pKa value is 13.0.

Clarification:

pKa + pKb = 14

For trimethylamine, Kb = 6.3 ×

Calculating pKb: pKb = - log (6.3 × )

= 1.0

Thus, pKa = 14 - pKb = 14 - 1.0

pKa = 13.0

Verification: The typical range for pKa in weak acids is from 2 to 13.

The visual representation is displayed in the following image.

For calculations, consider 100 grams of the compound:

ω(Cl) = 85.5% ÷ 100%.

ω(Cl) = 0.855; signifying the mass percentage of chlorine in the compound.

m(Cl) = 0.855 · 100 g.

m(Cl) = 85.5 g; this represents the mass of chlorine.

m(C) = 100 g - 85.5 g.

m(C) = 14.5 g; indicating the mass of carbon.

n(Cl) = m(Cl) ÷ M(Cl).

n(Cl) = 85.5 g ÷ 35.45 g/mol.

n(Cl) = 2.41 mol; this is the quantity of chlorine.

n(C) = 14.5 g ÷ 12 g/mol.

n(C) = 1.21 mol; this is the quantity of carbon.

n(Cl): n(C) = 2.41 mol: 1.21 mol = 2: 1.

The compound in question is identified as dichlorocarbene CCl₂.