207 is the mass number. 82 would be the atomic number
Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
Answer is: carbon.
<span>During gamma emission the nucleus emits radiation without
changing its composition, if for example have nucleus with six
protons and six neutrons (carbon atom) and after gamma decay there
is nucleus with six protons and six neutrons.
Gamma rays are the electromagnetic waves with
the shortest wavelengths (1 pm), highest frequencies (300 EHz) and
highest energy (1,24 MeV).</span>
Answer:
The precipitate was chromium hydroxide, which then reacted with more hydroxide to produce a soluble complex, Cr(OH)4
Explanation:
The following reaction takes place when chromium(III) nitrate reacts with NaOH:
+3 NaOH → 
(s)+ 
The precipitate that is formed is chromium hydroxide,
When more NaOH is added, the precipitate reacts with it which then results in the formation of a soluble complex ion:
(s) + 
(aq) → 
(aq)
is soluble complex ion
Hey there!
<span>Use the equation of Clapeyron:
</span>
T in kelvin :
26 + 273.15 => 299.15 K
R = 0.082
V = 10.2 L
P = 0.98 atm
number of moles :
P *V = n * R * T
0.98 * 10.2 = n * 0.082 * 299.15
9.996 = n * 24.5303
n = 9.996 / 24.5303
n = 0.4074 moles
Therefore:
Molar mass H2O = 18.01 g/mol
1 mole H2O ------------- 18.01 g
0.4074 moles ----------- m
m = 0.4074 * 18.01 / 1
m = 7.339 g of H2O
