<span><span>4.2×<span>1022</span></span><span>NA</span></span><span>
N_{A} is the avagadro number</span>
We are given with a 2.5 M stock solution of acetic acid and we are required to calculate the volume of the solution needed to prepare 100 milliliters of 0.5 M acetic acid solution. To solve this, we acquire the formula <span>Mconcentrated*Vconcentrated = Mdilute*Vdilute. That is 2.5 M*x=0.5M*100 ml where x is the volume of 2.5 M needed. x is equal to 20 ml. So we need 20 ml of 2.5 M solution and dilute to 100 ml using water as diluent.</span>
Aluminium reacts with dilute sulfuric acid based on the following reaction:
<span>2Al + 3H2SO4 ..............> Al2 (SO4)3 + 3H2
From the periodic table:
mass of aluminium = 27 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
mass of sulfur = 32 grams
Therefore:
molar mass of aluminium = 27 grams
molar mass of sulfuric acid = 2(1) + 32 + 4(16) = 98 grams
From the balanced chemical equation:
2 moles of aluminium react with 3 moles of dilute sulfuric acid.
This means that 34 grams of Al react with 294 grams of the acid
To get the amount of aluminium that reacts with </span><span>5.890 g of sulfuric acid, we will do cross multiplication as follows:
</span>amount of Al = (<span>5.890 x 34) / 294 = 0.6811 grams</span>
Answer:
67.4 % of C₉H₈O₄
Explanation:
To make titrations problems we know, that in the endpoint:
mmoles of acid = mmoles of base
mmoles = M . volume so:
mmoles of acid = 20.52 mL . 0.1121 M
mmoles of acid = mg of acid / PM (mg /mmoles)
Let's determine the PM of aspirin:
12.017 g/m . 9 + 1.00078 g/m . 8 + 15.9994 g/m . 4 = 180.1568 mg/mmol
mass (mg) = (20.52 mL . 0.1121 M) . 180.1568 mg/mmol
mass (mg) = 414.4 mg
We convert the mass to g → 414.4 mg . 1g / 1000mg = 0.4144 g
We determine the % → (0.4144 g / 0.615 g) . 100 = 67.4 %
Answer:
A general instrument, which is used to determine the concentration of hydrogen ion within the aqueous solution is known as a pH meter. The meter helps in determining the alkalinity or acidity, which is articulated in the form of pH. It is also called a potentiometric pH meter as it helps in finding the variation in electrical potential between a reference electrode and a pH electrode. This electrical potential variation is associated with the pH of the solution.
The potentiometric pH meter comprises a pair of electrodes and a basic electronic amplifier, some may even comprise a combination electrode and some sort of display that demonstrates pH units. The potentiometric pH meter generally exhibits a reference electrode or a combination electrode, and a glass electrode. The probes or electrodes are administered within a solution whose pH values are needed to be determined.