<h2>The answers are
and
</h2>
Explanation:
Given -
a) The molecular formula of ethylene glycol -
∴ The empirical formula of ethylene glycol will be -
Given -
b) The molecular formula of per-oxo-disulfuric acid (a compound used in bleaching agents) -
∴ The empirical formula of per-oxo-disulfuric acid will be -
Hence, the answers are and .
We calculate for the amount of chromium metal in the reactant by,
= 350 x (mass of Cr2/mass of Cr2O3)
= 350 x (104/152)
= 239.47 grams
The amount of Cr metal in the product is only 213.2 grams. Thus, the percent yield.
percent yield = (213.2 grams/239.47) x 100%
= 89%
<span>Melting is an endothermic process (i.e. it absorbs heat), whereas freezing is an exothermic process (i.e. it releases heat, or can be thought of, albeit incorrectly from a thermodynamics standpoint, as "absorbing cold"). The standard enthalpy of fusion of water can be used for both scenarios, but standard enthalpy is in units of energy/mass, so 10 times as much energy will be absorbed in the former scenario (melting 10 kg of ice) than what will be absorbed in the latter scenario (freezing 1 kg of water). For both processes, assuming the water is pure and at standard atmospheric pressure, and the entire mass remains at thermal equilibrium, the temperature of both the solid and the liquid will remain at precisely 0 degrees Celsius (273 K) for the duration of the phase change.</span>
Answer:
A titration
Explanation:
A common example of a titration is when we have an acid of unknown concentration, so we add a known volume of a base of known concentration. This process lets us determine the concentration of the acid.
By definition, a titration is a quantitative analysis, as we determine how much of an analyte is there in a sample. However, <u>there are quantitative analyzes which are not titrations</u>. This is why the most appropiate answer is<em> a titration</em>.
Answer is: 6.022·10²² molecules of glucose.
c(glucose) = 100 mM.
c(glucose) = 100 · 10⁻³ mol/L.
c(glucose) = 0.1 mol/L; concentration of glucose solution.
V(glucose) = 1 L; volume of glucose solution.
n(glucose) = c(glucose) · V(glucose).
n(glucose) = 0.1 mol/L · 1 L.
n(glucose) = 0.1 mol; amount of substance.
N(glucose) = n(glucose) · Na (Avogadro constant).
N(glucose) = 0.1 mol · 6.022·10²³ 1/mol.
N(glucose) = 6.022·10²².
