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  1. ... specific heat of ice is 2.06 kJ/kgC specific heat of steam is 2.1 kJ/kgK heat of fusion of ice is 334 kJ...

    1 Answers · Science & Mathematics · 20/12/2017

  2. ...x (863 g) x (123 - 100)°C = 39500 J = 39.500 kJ to heat the steam to 123°C. 52.557 kJ + 288.380 kJ + 360.734 kJ + 1953.992...

    2 Answers · Science & Mathematics · 19/12/2017

  3. (4.184 J/(g ∙ K)) x (430 g) x (100 - 52)K = 86358 J to warm the water to its boiling point (2259.4 J/g) x (430 g) = 971542 J to vaporize the water (2.03 J/(g*K)) x (430 g) x (120 - 100)K = 17458 J to heat the vapor to 120℃ 86358 J...

    2 Answers · Science & Mathematics · 13/12/2017

  4. 1.45 kJ/0.652 grams H2O x 18.0 grams water/mole H2O = 40.0 kilojoules/mole H2O

    1 Answers · Science & Mathematics · 19/01/2018

  5. ... 0°C or 100°C, it seems safe to assume that all the steam condensed and all the ice melted. (540 cal/g) x (4 g) = 2160 cal lost...

    1 Answers · Science & Mathematics · 28/11/2017

  6. Calculate moles of water in the sample. Molar mass NiSO4.6H2O = 262.85 g/mol 2.80 g / 262.85 g/mol X (6 mol H2O/1 mol NiSO4.6H2O) = 0.0639 mol H2O Now use ideal gas law to calculatte volume: PV = n RT 1 atm...

    1 Answers · Science & Mathematics · 26/11/2017

  7. A. (2.087 J/(g·°C)) x (57.0 g) x (0 − (−18.0))°C = 2141.3 J to warm the ice to its melting point (333.6 J/g) x (57.0 g) = 19015.2 J to melt the ice (4.184 J/(g·°C)) x (57.0 g) x (25.0 - 0)°C = 5962.2 J to warm the melted ice...

    1 Answers · Science & Mathematics · 21/11/2017

  8. ...08205746 L atm/K mol) x (100 + 273) K / (1.00 atm)) = 41.7 L steam

    1 Answers · Science & Mathematics · 20/11/2017

  9. ...g / 18.01 g/mol) = 30.433 kJ = 30433 J Heat to raise temp of steam from 100 to 110 C: q = m c (T2-T1) q = 13.5 g (2.03 J...

    1 Answers · Science & Mathematics · 19/11/2017

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    1 Answers · Science & Mathematics · 17/11/2017

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