Because mass is not always the same for objects made from the same material. Just because an object has the same mass as another doesn’t mean they are the same material.
Millimeters are an SI unit of length that =0.0001 m
Answer:
A. 0.143 M
B. 0.0523 M
Explanation:
A.
Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).
KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄
The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:
1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol
The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.
5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:
M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M
B.
Let's consider the neutralization of potassium hydroxide and perchloric acid.
KOH + HClO₄ → KClO₄ + H₂O
When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.
Answer:
b. 2.28 M
Explanation:
The reaction of neutralization of NaOH with H2SO4 is:
2NaOH + H2SO4 → Na2SO4 + 2H2O
<em>Where 2 moles of NaOH react per mole of H2SO4</em>
<em />
To solve the concentration of NaOH we need to find the moles of H2SO4. Using the chemical equation we can find the moles of NaOH that react and with the volume the molar concentration as follows:
<em>Moles H2SO4:</em>
45.7mL = 0.0457L * (0.500mol/L) = 0.02285 moles H2SO4
<em>Moles NaOH:</em>
0.02285 moles H2SO4 * (2moles NaOH / 1 mol H2SO4) = 0.0457moles NaOH
<em>Molarity NaOH:</em>
0.0457moles NaOH / 0.020L =
2.28M
Right option:
<h3>b. 2.28 M</h3>
Curium (Cm, 96) – Pierre and Marie Curie einsteinium (Es, 99) – Albert Einsteinfermium (Fm, 100) – Enrico Fermigallium (Ga, 31) – both named after Gallia (Latin for France) and its discoverer, Lecoq de Boisbaudran (le coq, the French word for 'rooster' translates to gallus in Latin)hahnium (105) – Otto Hahn (Dubnium, named for Dubna in Russia, is the IUPAC-accepted name for element 105)lawrencium (Lr, 103) – Ernest Lawrencemeitnerium (Mt, 109) – Lise Meitner<span>mendelevium (Md, 101) – Dmitri Mende</span>
<span>obelium (No, 102) – Alfred Nobel<span>roentgenium (Rg, 111) – Wilhelm Roentgen (formerly Ununumium)</span><span>rutherfordium (Rf, 104) – Ernest Rutherford </span><span>seaborgium (Sg, 106) – Glenn T. Seaborg</span></span>
