Answer:
See explanation
Explanation:
A. This is a neutralization reaction.
Molecular equation;
HBr(aq) + CsOH(aq) ---------> CsBr(aq) + H20(l)
Complete ionic equation;
H^+(aq) + Br^-(aq) + Cs^(aq) + OH^-(aq) --------> Cs^+(aq) + Br^- + H20(l)
Net ionic equation;
H^+(aq) + OH^-(aq) --------> H20(l)
B. This is a gas forming reaction;
H2SO4(aq) + Na2CO3(aq) ------->Na2SO4(aq) + H2O(l) + CO2(g)
Complete ionic equation;
2H^+(aq) + SO4^-(aq) + 2Na^+(aq) + CO3^2-(aq) ------->2Na^+(aq) + SO4^-(aq) + H2O(l) + CO2(g)
Net ionic equation;
2H^+(aq) + CO3^2-(aq) -------> + H2O(l) + CO2(g)
C. This a precipitation reaction
Molecular equation;
CdCl2(aq) + Na2S(aq) ------->CdS(s) + 2NaCl(aq)
Complete ionic equation;
Cd^2+(aq) + 2Cl^-(aq) + 2Na^+(aq) + S^2-(aq) ---------> CdS(s) + 2Na^+(aq) + 2Cl^-(aq)
Net ionic equation;
Cd^2+(aq) + S^2-(aq) ---------> CdS(s)
Answer:
15.89%
Explanation:
To calculate the percentage of carbon in BeC2O4•3H2O, first we calculate the molar mass of BeC2O4•3H2O
MM of BeC2O4•3H2O = 9+(12x2)+(4x16)+3(2+16) = 9+24+64+54 =151g/mol
Mass of C in BeC2O4•3H2O = 2x12 = 24g
%Mass of carbon in BeC2O4•3H2O = (24/151) x 100
= 15.89%
Answer:
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