Could you please explain in detail how to do these questions? Thank you in advance.
1 ) we have to consider number of ions x concentration, higher the number higher will be conductivity.
( a ) 3 x 2 = 6
( b ) 2.4 x 3 = 7.2
( c ) 3.2 x 2 = 6.4
( d ) 2.1 x 4 = 8.4
correct answer is ( d )
2 ) answer is ( d ) since K2SO4 does not form precipitate.
3 ) balance reaction is option ( c ) .it is neutralisation reaction.
4 )
Could you please explain in detail how to do these questions? Thank you in advance. ---...
Could you please explain in detail how to do these questions? Thank you in advance. 11. (8 points) Fill in the blanks to complete each of the following nuclear reactions. - - 233 Ac + He 2. Ca + - - 1K c. 20. Au → 291 Hg + d. 18; Eu → __ thiB 12. (10 points) The molar nuclear mass of boron-10 is 10.012937 g/mol. The molar mass of a proton is 1.007825 g/mol. The molar mass of...
325 mL of a 0.0700M solution of sodium hydroxide reacts with 75.0 mL of a 0.115 M solution of cobalt (III) sulfate. What mass (in g) of cobalt (III) hydroxide( MM = 109.96) is formed assuming the reaction went to completion? NaOH(aq) + Co2(SO4)3(aq) --> Na2SO4(aq) + Co(OH)3(s) (unbalanced) Write answer to three significant figures.
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Could someone answer the following questions? 1) 11.2 g of potassium hydroxide (KOH) is dissolved in sufficient water to make 1 L of solution. What is the concentration of KOH in the solution? a. 0.02 mol/L b. 0.1 mol/L c. 0.2 mol/L d. 0.01 mol/L 2) If the solution is neutral, which of the following must be true? a. [OH-] = [H2O] b. [H+] > [OH-] c. [H+] < [OH-] d. [H+] = [OH-] 3)A student titrates 0.5222 grams of...
325 mL of a 0.0700M solution of sodium hydroxide reacts with 75,0 mL of a 0.115 M solution of cobalt(III) sulfate. What mass (in g) of cobalt(III) hydroxide( MM = 109.96) is formed assuming the reaction went to completion? NaOH(aq) + CO2(SO4)3(aq) --> Na2SO4(aq) + C(OH)3(s) (unbalanced) Write answer to three significant figures. (8 00:40:48 Numeric Response
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3) How many NH3 molecules are produced when 8.00 mol of Ca(OH)2 are completely reacted according to: (Avogadro's number = 6.022 x 1023 entities"/mol) (NH4)2SO4(aq) + Ca(OH)2(aq) → 2NH3(g) + Caso (aq) + 2H,00) (2pts)