Answer:
4.43L is final volume of the ballon
Explanation:
Avogadro's law of ideal gases states that <em>equal volumes of gases, at the same temperature and pressure, have the same number of molecules</em>.
The formula is:
Where V and n are volume and moles of the gas in initial and final conditions.
If the initial conditions are 0.0145 moles and 2.54L and final amount of moles is 0.0253moles, final volume is:
V₂ = <em>4.43L is final volume of the ballon</em>
Answer:
12 %
Explanation:
Produces 15 J of work for input 125 J
15/125 * 100% = 12%
Answer:
Atoms consist of three basic particles: protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged).
Explanation:
Answer:
2.25g of NaF are needed to prepare the buffer of pH = 3.2
Explanation:
The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:
pH = pKa + log [A-] / [HA]
<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>
The moles of HF are:
500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF
Replacing:
3.2 = 3.17 + log [A-] / [0.0500moles]
0.03 = log [A-] / [0.0500moles]
1.017152 = [A-] / [0.0500moles]
[A-] = 0.0500mol * 1.017152
[A-] = 0.0536 moles NaF
The mass could be obtained using the molar mass of NaF (41.99g/mol):
0.0536 moles NaF * (41.99g/mol) =
<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>
The correct answer is approximately 11.73 grams of sulfuric acid.
The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.
The balanced equation is:
2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O
Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g
40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.
