Much heat (in calories) must be applied to a 100 ml beaker filled with boiling water in order to vaporize the water is: 540 calories needed per gm of water of turn to steam, so to boling 100 ml water you need 54,000 (5.4 x 10^4) calories into steam.
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Answer is: <span>the molarity of this glucose solution is 0.278 M.
m</span>(C₆H₁₂O₆<span>) = 5.10 g.
n</span>(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆<span>) .
</span>n(C₆H₁₂O₆) = 5.10 g ÷ 180.156 g/mol.
n(C₆H₁₂O₆<span>) = 0.028 mol.
</span>V(solution) = 100.5 mL ÷ 1000 mL/L.
V(solution) = 0.1005 L.
c(C₆H₁₂O₆) = n(C₆H₁₂O₆) ÷ V(solution).
c(C₆H₁₂O₆) = 0.028 mol ÷ 0.1005 L.
c(C₆H₁₂O₆<span>) = 0.278 mol/L.</span>
Answer: Explanation:
Gallium has three electrons in the outer energy level and therefore has three valence electrons. The identification of valence electrons is vital because the chemical behavior of an element is determined primarily by the arrangement of the electrons in the valence shell.
Answer:
-0.044 V
Explanation:
The overall electrode potential is usually estimated by subtracting the potential of the reference electrode from the potential of the system in question. In the given problem, the standard electrode potential of silver-silver chloride with respect to the standard hydrogen electrode is +0.197 V. In addition, the potential of the calomel reference electrode with respect to the standard hydrogen electrode is +0.241 V. Using the data above, the potential of with respect to silver-silver chloride is equivalent to 0.197-0.241 = -0.044 V.
It takes 21.3 days
<h3>Further explanation</h3>
Given
5 hr = 8 kg Alcohol
Required
Days to consume 1000 kg of glucose
Solution
Alcoholic fermentation⇒ glucose produces ethanol and carbon dioxide,
C₆H₁₂O₆ → 2 C₂H₅OH + 2CO₂
mol ethanol :
moles of glucose to produce 108.7 moles ethanol :
54.35 moles = 5 hours
moles of 1000 kg of glucose :
So for 5555.5 moles, it takes :