Answer:
The answer to your question is: 3.745 x 10 ²³ atoms
Explanation:
Data
density = 8.96 g/cm³
mass = 29.54 g
atomic radius = 1.28 A° 1.28 x 10⁻¹⁰ m
Volume of a cube
density = mass / volume
volume = mass / density
volume = 29.54 / 8.96
volume = 3.29 cm³
Volume of a copper atom
volume = (4/3) πr³
volume = (4/3) π (1.28 x 10 ⁻⁸)³
= 8.784 x 10 ⁻²⁴ cm³
Number of atoms = 3.29 cm³ / 8.784 x 10 ⁻²⁴ cm³
= 3.745 x 10 ²³ atoms
Answer: A
Explanation: the only way to turn it from blue to yellow is to mix it with an acidic solution.
Answer:
1.4 × 10^-4 M
Explanation:
The balanced redox reaction equation is shown below;
5Fe2+ + MnO4- + 8H+ --> 5Fe3+ +Mn2+ + 4H2O
Molar mass of FeSO4(NH4)2SO4*6H2O = 392 g/mol
Number of moles Fe^2+ in FeSO4(NH4)2SO4*6H2O = 3.47g/392g/mol = 8.85 × 10^-5 moles
Concentration of Fe^2+ = 8.85 × 10^-5 moles × 1000/200 = 4.425 × 10^-4 M
Let CA be concentration of Fe^2+ = 4.425 × 10^-4 M
Volume of Fe^2+ (VA)= 20.0 ml
Let the concentration of MnO4^- be CB (the unknown)
Volume of the MnO4^- (VB) = 12.6 ml
Let the number of moles of Fe^2+ be NA= 5 moles
Let the number of moles of MnO4^- be NB = 1 mole
From;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CB= CAVANB/VBNA
CB= 4.425 × 10^-4 × 20 × 1/12.6 × 5
CB = 1.4 × 10^-4 M
Answer:
1.97 moles of Na contains greater number of atoms than 6.8 × 10²² atoms of Li.
Explanation:
To solve this problem we will first calculate the number of atoms contained by 1.97 moles of Na and then will compare it with 6.8 × 10²² atoms of Li.
As we know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of atoms contained by 1.97 moles of Na, we will use following relation,
Moles = Number of Atoms ÷ 6.022 × 10²³ atoms.mol⁻¹
Solving for Number of Atoms,
Number of Atoms = Moles × 6.022 × 10²³ Atoms.mol⁻¹
Putting values,
Number of Atoms = 1.07 mol × 6.022 × 10²³ Atoms.mol⁻¹
Number of Atoms = 1.18 × 10²⁴ Atoms of Na
Hence,
1.97 moles of Na contains greater number of atoms than 6.8 × 10²² atoms of Li.
They are of the same period and are similar in some form
