for 39g water solute dissolved at 20C = solubility ( g/ 100 g H2O ) × mass of water = ( 11g / 100g H2O ) × 39g H2O = 4.29 g
amount of solute dissolved at 30 C =
= 23 / 100 * 39 = 8.97 g
Amount of extra solute dissolved = 8.97 - 4.29 = 4.7 g
Answer is: volume of carbon dioxide is 1,84·10⁸ l.
Chemical reaction: C + O₂ → CO₂.
m(C) = 100 t · 1000 kg/t = 100000 kg
m(C) = 100000 kg · 1000 g/kg = 10⁸ g.
n(C) = m(C) ÷ M(C).
n(C) = 10⁸ g ÷ 12 g/mol.
n(C) = 8,33·10⁶ mol.
From chemical reaction: n(C) . n(CO₂) = 1 : 1.
n(CO₂) = 8,33·10⁶ mol.
m(CO₂) = 8,33·10⁶ mol · 44 g/mol.
m(CO₂) = 3,66·10⁸ = 3,66·10⁵ kg.
V(CO₂) = 3,66·10⁵ kg ÷ 1,98 kg/m³ = 1,84·10⁵ m³.
V(CO₂) = 1,84·10⁵ m³ · 1000 l/m³ = 1,84·10⁸ l.
Answer:
it is s because enegy allows us to walk
To answer this problem, we use Hess' Law to calculate the overall enthalpy of the reactions. The goal is to add all the reactions such that the final reaction is C<span>5H12 (g) + 8O2 (g) → 5CO2 (g) + 6H2O (l) through cancellation adn multiplication. The first equation is multiplied by 5, the second one is multiplied by 6 and the third one is reversed. The final answer is -3538 J or -3.54 x10^3 kJ.</span>
Answer:
Mercury
Explanation:
While it is true that most metals are solid at room temperature, mercury is liquid at room temperature hence mercury is often designated as the 'liquid metal'.
Thus, if i find a bottle on the shelf that has no solid in it, only liquid and i know that only pure metals are stored in that area of the laboratory, then i will quickly relabel it as mercury.
