For stainless steel different kinds of compositions are used. Based on that different series of stainless steel has been coined.
1. Series 200 - Iron alloyed with <span>chromium, nickel and manganese.
2. Series 300 - It has
a. Stainless Steel 304 - it has composition of 18% chromium and 8% Nickel
b. </span>Stainless Steel 316 - This has 18% chromium and 10% Nickel
Each kind of stainless steel is of different cost and has different applications.
Answer:
This question is incomplete, here's the complete question:
<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>
Explanation:
Reaction :-
K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4
Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol
Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol
Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L
Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L
Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol
Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.
0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+
Final concentration of potassium cation
= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M
The answer to this question would be: 3.125%
Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
Answer:
A. H3O+/OH−
Explanation:
A conjugate acid-base pair are a pair of molecules that differ in 1 H⁺
A. The conjugate pair of H₃O⁺ = H₂O not OH⁻
B. The conjugate pair of NH₄⁺ is NH₃
C. The conjugate pair of C₂H₃O₂⁻ is HC₂H₃O₂
D. The conjugate pair of H₂SO₃ is HSO₃⁻
That means right option, that is not a conjugate acid-base pair, is:
<h3>A. H3O+/OH−
</h3>
Answer:
KI
Explanation:
From the question, we can see that a qualitative analysis of the compound shows that it has a lilac flame colour. The lilac flame colour corresponds to the potassium ion (K^+).
Again, the test of addition of HNO3(aq) and AgNO3(aq) to a solution is a test for halogens. If the result is a green precipitate, then the ion present is the iodide ion (I^-).
Hence, the compound must be KI.
