Answer:
Q = 1267720 J
Explanation:
∴ QH2O = mCpΔT
∴ m H2O = 500 g
∴ Cp H2O = 4.186 J/g°C = 4.183 E-3 KJ/g°C
∴ ΔT = 120 - 50 = 70°C
⇒ QH2O = (500 g)(4.183 E-3 KJ/g°C)(70°C) = 146.51 KJ
∴ ΔHv H2O = 40.7 KJ/mol
moles H2O:
∴ mm H2O = 18.015 g/mol
⇒ moles H2O = (500 g)(mol/18.015 g) = 27.548 mol H2O
⇒ ΔHv H2O = (40.7 KJ/mol)(27.548 mol) = 1121.21 KJ
⇒ Qt = 146.51 KJ + 1121.21 KJ = 1267.72 KJ = 1267720 J
Before an article is published in a scientific journal or in any "peer-reviewed" journal the article is reviewed thoroughly by scholars from the journal as well as peers or scholars of the articles author from the same field. This process occurs to provide credibility to the ideas being published and so that readers and other scholars can rely on the validity of the material being published.
Answer:
0.00013
Explanation:
I'm pretty sure. I first subtracted 1.87 from 2. I got 0.13. The conversion from mL to L is 0.001. I multiplied 0.13 and 0.001 and got <u>0.00013 mL.</u>
Answer:
they get colder and darker, with less light
Explanation:
Answer:
Na₂₆F₁₁
Explanation:
We find the moles of the substance assuming 100 g of the substance is present. Why do we take 100 g? Because then the percent of sodium/fluorine, would be the g of sodium/fluorine respectively:
74.186 g Sodium | 1 mol Sodium/23 g => 3.2255 mol Na
25.814 g Fluorine | 1 mol Fluorine/19 g => 1.3586 mol F
Divide each by smallest number of moles:
3.2255/1.3586 = 2.37
1.3586/1.3586 = 1
Multiply by common number to get a smallest whole number:
2.37*11 = 26,
1*11 = 11
The empirical formula is Na₂₆F₁₁
