Firstly, calculate the molar using the following equation M = 10XdensityXpercentage(37 in your case)/Molecular Weight. Then calculate the required molarity by dilution.
1 Answers 1 views∆H = m x s x ∆T-30.9 = 4.2 × 4 × ∆T-30.9 = 16.8 × ∆T∆T = -30.9/16.8 = 1.84°C
1 Answers 1 views150 mL
1 Answers 1 views2K + H2SO4 = K2SO4 + H2 if m(H2) = 6 g, so n(H2) = 6 g / 2 g/mol = 3 moles; n(H2) : n(K2SO4) = 1 : 1;...
1 Answers 1 viewsSolution:The balanced chemical equation:2KBr + Cl2 → 2KCl + Br2According to the equation above: n(Br2) = n(KBr)/2Moles of KBr = Mass of KBr / Molar mass of KBrThe molar mass of...
1 Answers 1 viewsCaCO3 = CaO + CO2 nCO2 = nCaCO3 = mCaCO3/MWCaCO3 = 10.5 g/100.0869 g/mol = 0.1049 mol; VCO2 = nCO2*Vm = 0.1049 mol*24 L/mol = 2.5178 L ≈ 2.52 L
1 Answers 1 viewsCO2 + Ca(OH)2 = CaCO3 + H2Om(CaCO3) = 10.0 g(10 g CaCO3 / 1)(1 mol / 100.1 g)(1 mol CO2 / 1 mol CaCO3)(22.4 L STP / 1 mol) = 2.24 L CO2 STP
1 Answers 1 views535 g
1 Answers 1 viewsNo.Combustion reaction - ecsotermical prosecc, with intensiv burn energy (May be hv)this would be called neutralisation, if you take equivival ammount. Also or acid-base reaction
1 Answers 1 viewspV=nRT. n=pV/RT = 695 x 4500 / 62400 x 708= 0,0707912321 moles.0,0707912321 x 17 = 1,20345095 g.
1 Answers 1 views