Highest percentage of gold by weight can be determined by looking at the atomic mass of the other elements in the compound. the smaller the total of mass of the other elements, the heigher the percentage of Au will be in that compound
a. AuOH, OH- = 17 gram/mole
b. Au(OH)3, 3 x OH- = 3 x 17 = 51 gram/mole
c. AuCl3 = 3x35.5= 106.5 gram/mole
d. AuI3 = 3x127 = 381 gram/mole
as we can see AuOH has the least Molecular mass, hence % of Au will be the highest compared to other choices
Answer: A
The main alloying constituents are copper, manganese, and magnesium. A commonly used modern equivalent of this alloy type is AA2024, which contains 4.4% copper, 1.5% magnesium, 0.6% manganese and 93.5% aluminium by weight. Typical yield strength is 450 MPa
<span>If you do the math you can figure out the chemical formula. Just use the empirical formula
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Answer: 2.8275grams
Explanation: A buffer is made btw a weak acid and it salt. In a solution made by dissolving a weak acid in solution, equilibrium is set up btw ionised and unionised ion. For Benzoic acid
C6H5COOH....> C6H5COO- + H+
Ka = [C6H5COO-] [H+]/ [C6H5COOH].......(1)
using Ka = 6.5× 10^-5, [C6H5COOH] = 0.02M. PH= - log[H+] ....> [H+]= 10^-4M.
Putting the values in(1)
[C6H5COO-]= 6.5× 10^-5 × 0.02/ 10^-4
[C6H5COO-] = 0.013M = Molarity of sodium benzoate
Mole(C6H5COONa) = 0.013 × Volume = 0.013mol/litre × 1.5 litre
Mole(C6H5COONa) = 0.0195mol
Mass(C6H5COONa) = 0.0195 × Molar mass
Mass(C6H5COONa) = 2.8275g
Answer:
When a sodium atom transfers an electron to a chlorine atom, forming a sodium cation (Na+) and a chloride anion (Cl-), both ions have complete valence shells, and are energetically more stable. The reaction is extremely exothermic, producing a bright yellow light and a great deal of heat energy.
Answer:
Empirical formula is PNCl₂
Explanation:
Percent composition means that 100 g of compound has x g of each element.
In 100 g of compound x, we have 26.73 g of P, 12.09 g of N, 61.18 g of Cl.
So, let's make some rules of three:
In 100 g of compound we have 26.73 g of P, 12.09 g of N, 61.18 g of Cl
In 579.43 g of compound we have:
(579.43 . 26.73) / 100 = 155 g of P
(579.43 . 12.09) / 100 = 70 g of N
(579.43 . 61.18) / 100 = 354 g of Cl
Let's convert the mass of the elements in moles.
155 g of P / 30.97 g/mol = 5 P
70 g / 14 g/mol = 5 N
354 g / 35.45 g/mol = 10 Cl