Answer:
NH3(aq)
Explanation:
Gold III hydroxide is an inorganic compound also known as auric acid. It can be dehydrated at about 140°C to yield gold III oxide. Gold III hydroxide is found to form precipitates in alkaline solutions hence it is not soluble in calcium hydroxide.
However, gold III hydroxide forms an inorganic complex with ammonia which makes the insoluble gold III hydroxide to dissolve in ammonia solution. The equation of this complex formation is shown below;
Au(OH)3(s) + 4 NH3(aq) -------> [Au(NH3)4]^3+(aq) + 3OH^-(aq)
Hence the formation of a tetra amine complex of gold III will lead to the dissolution of gold III hydroxide solid in aqueous ammonia.
Answer is: <span>number of grams of oxygen present in a sample is </span>37,41 g.
m(S) = 50 g.
mass percentage (S) = 57,20 % = 0,572.
mass percentage (O) = 100% - 57,20%
mass percentage(O) = 42,8% = 0,428.
m(O) = ?.
make proportion 57,20% : 42,8% = 50 g : m(O)
57,2% · m(O) = 42,8% · 50 g.
m(O) = 42,8% · 50 g ÷ 57,2%.
m(O) = 37,41 g.
Answer is: Protons and neutrons are located in the nucleus.
Atomic number (Z) is total number of protons and mass number (A) is total number of protons and neutrons in a nucleus.
Nucleus of an atom has positive charge, electrons are negative and atom has neutral net charge.
Helium (symbol: He) is an element with atomic number 2 (2 protons) and mass number 4 (2 neutrons; n° = 4 - 2 = 2).
CH3CH2CH2CH3 < CH3CH2CHO < CH3CHOHCH3
Explanation:
Boiling point trend of Butane, Propan-1-ol and Propanal.
Butane is a member of the CnH2n+2 homologous series is an alkane. Alkanes have C-H and C-C bonds which have Van der waals dispersion forces which are temporary dipole-dipole forces (forces caused by the electron movement in a corner of the atom). This bond is weak but increases as the carbon chain/molecule increases.
In Propan-1-ol(Primaryalcohol), there is a hydrogen bond present in the -OH group. Hydrogen bond is caused by the attraction of hydrogen to a highly electronegative element like Cl-, O- etc. This bond is stronger than dispersion forces because of the relative energy required to break the hydrogen bond. Alcohols (CnH2n+1OH) also experience van der waals dispersion forces on its C-C chain and C-H so as the Carbon chain increases the boiling point increases in the homologous series.
Propanal which is an Aldehyde (Alkanal) with the general formula CnH2n+1CHO. This molecule has a C-O, C-C and C-H bonds only. If you notice, the Oxygen is not bonded to the Hydrogen so there is no hydrogen bond but the C-O bond has a permanent dipole-dipole force caused by the electronegativity of oxygen which is bonded to carbon. It also has van der waals dispersion forces caused by the C-C and C-H as the carbon chain increases down the homologous series. The permanent dipole-dipole forces are not as easy to break as van der waals forces.
In conclusion, the hydrogen bonds present in alcohols are stronger than the permanent dipole-dipole bonds in the aldehyde and the van der waals forces in alkanes (irrespective of the carbon chain in Butane). So Butane < Propanal < Propan-1-ol
First, find percent of oxygen: atom/molecule... there are 2 atoms of Oxygen so: O2/C2H2O2 which is: 32g O2 / 58g C2H2O2 =32/58.
<span>Next, multiply this by the total mass (56g C2H2O2) and the units will cancel out (g*g/g -> g) leaving you with the mass of Oxygen: </span>
<span>56g C2H2O2 * 32g O2/58g C2H2O2 = 56*32/58= 31g</span>
