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>
Answer:
The half-life of a radioisotope describes the amount of time it takes for said isotope to decay to one-half the original amount present in the sample.
Nitrogen-13, because it has a half-life of ten minutes, will experience two half-lives over the course of the twenty minute period. This means that 25% of the isotope will remain after this.
0.25 x 128mg = 32mg
32mg of Nitrogen-13 will remain after 20 minutes.
Answer:
Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + 2KNO₃
Step-by-step explanation:
The unbalanced equation is
Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + KNO₃
Notice that the complex groups like NO₃ and CrO₄ stay the same on each side of the equation.
One way to simplify the balancing is to replace them with a single letter.
(a) For example, let <em>X = NO₃</em> and <em>Y =CrO₄</em>. Then, the equation becomes
PbX₂ + K₂Y ⟶ PbY + KX
(b) You need 2X on the right, so put a 2 in front of KX.
PbX₂ + K₂Y ⟶ PbY + 2KX
(c) Everything is balanced. Now, replace X and Y with their original meanings. The balanced equation is
Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + 2KNO₃
Answer:
Yes it is
Explanation: because it is composed of identical molecules consisting of atoms of two or more chemical elements.
hope it helps <3
The balanced equation for the above reaction is as follows;
Na₂SO₄ + BaCl₂ --> BaSO₄ + 2NaCl
Na₂SO₄ reacts with BaCl₂ in the molar ratio 1:1
Number of Na₂SO₄ moles - 10.0 g / 142.1 g/mol = 0.0704 mol
Number of BaCl₂ moles - 10.0 g / 208.2 g/mol = 0.0480 mol
this means that 0.0480 mol of each reactant is used up, BaCl₂ is the limiting reactant and Na₂SO₄ has been provided in excess.
stoichiometry of BaCl₂ to BaSO₄ is 1:1
number of BaSO₄ moles formed - 0.0480 mol
Mass of BaSO₄ - 0.0480 mol x 233.2 g/mol = 11.2 g
theoretical yield is 11.2 g but the actual yield is 12.0 g
the actual product maybe more than the theoretical yield of the product as the measured mass of the actual yield might contain impurities.
percent yield - 12.0 g/ 11.2 g x 100% = 107%
this is due to impurities present in the product or product could be wet.
