Answer:
1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.
Explanation:
Given data:
mass of cadmium = 6.35 g
Number of atoms of aluminum as 6.35 g cadmium contain = ?
Solution:
Number of moles of cadmium = 6.35 g/ 112.4 g/mol
Number of moles of cadmium = 0.06 mol
Number of atoms of cadmium:
1 mole = 6.022×10²³ atoms of cadmium
0.06 mol × 6.022×10²³ atoms of cadmium/ 1mol
0.36×10²³ atoms of cadmium
Number of atoms of Al:
Number of atoms of Al = 0.36×10²³ atoms
1 mole = 6.022×10²³ atoms
0.36×10²³ atoms × 1 mol /6.022×10²³ atoms
0.06 moles
Mass of aluminum:
Number of moles = mass/molar mass
0.06 mol = m/ 27 g/mol
m = 0.06 mol ×27 g/mol
m = 1.62 g
Thus, 1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.
Answer:
c. 3
Explanation:
The unbalanced reaction expression is given as:
Fe + O₂ → Fe₂O₃
To balance the expression gives the coefficient of oxygen.
Assign the coefficients a,b, c and use a mathematical approach to solve this problem:
aFe + bO₂ → cFe₂O₃
Conserving Fe: a = 2c
O: 2b = 3c
Now let c = 1, a = 2, b =
Multiply through by 2,
a = 4, b = 3 and c = 2
4Fe + 3O₂ → 2Fe₂O₃
The coefficient of O₂ is 3
Answer:
2
Explanation:
your answer is correct. The overall balanced equation would be:
2Cu +2HNO3 ----> 2CuNO3 +H2
Answer:
Semiconductors are poor conductors at low temperatures, but their resistance decreases with increasing temperature.
Explanation:
A semiconductor can be defined as a crystalline solid substance that has its conductivity lying between that of a metal and an insulator, due to the effects of temperature or an addition of an impurity. Semiconductors are classified into two main categories;
1. Extrinsic semiconductor.
2. Intrinsic semiconductor.
The statement which best describes the electrical conductivity of metals and semiconductors is that semiconductors are poor conductors at low temperatures, but their resistance decreases with increasing temperature.
This ultimately implies that, semiconductors are typically an insulator (poor conductor) at low temperatures and a good conductor at high temperatures.
Additionally, conduction involves the transfer of electric charge or thermal energy due to the movement of particles. When the conduction relates to electric charge, it is known as electrical conduction while when it relates to thermal energy, it is known as heat conduction.
Answer:
Ksp = [ Cu+² ] [ OH-] ²
molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol
Ksp = [ Cu+² ] [ OH-] ²
Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰
|__________|___<u>Cu</u><u>+</u><u>²</u><u> </u>__|_<u>2</u><u>OH</u><u>-</u>____|
|<u>Initial concentration(M</u>)|___<u>0</u>__|_<u>0</u>______|
<u>|Change in concentration(M)</u>|_<u>+S</u><u> </u>|__<u>+2S</u>__|
|<u>Equilibrium concentration(M)|</u><u>_S</u><u> </u><u>_</u><u>|</u><u>2S___</u><u>|</u>
Ksp = [ Cu+² ] [ OH-] ²
2.2 ×10-²⁰ = (S)(2S)²= 4S³
![s = \sqrt[3]{ \frac{2.2 \times {10}^{ - 20} }{4} } = 1.8 \times {10}^{ - 7}](https://tex.z-dn.net/?f=s%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B2.2%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%2020%7D%20%7D%7B4%7D%20%7D%20%20%3D%201.8%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%207%7D%20)
S = 1.8 × 10-⁷ M
The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M
Solubility of Cu (OH)2 =

<h3>
Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>
I hope I helped you^_^