Answer:
C Group 11
Explanation:
Group 11, by modern IUPAC numbering, is a group of chemical elements in the periodic table, consisting of copper (Cu), silver (Ag), and gold (Au).
These elements show highest electrical conductivity.
Answer:
D) 0.50 mole of Ne
Explanation:
Given data:
Number of molecules of nitrogen = 3.0×10²³ molecules
Which sample contain same number of molecules as nitrogen= ?
Solution:
A) 0.25 mole of O₂
1 mole = 6.022×10²³ molecules
0.25 mol × 6.022×10²³ molecules / 1 mol
1.51×10²³ molecules
B) 2.0 moles of He.
1 mole = 6.022×10²³ molecules
2.0 mol × 6.022×10²³ molecules / 1 mol
12.044×10²³ molecules
C) 1.0 moles of H₂
1 mole = 6.022×10²³ molecules
D) 0.50 mole of Ne
1 mole = 6.022×10²³ molecules
0.50 mol × 6.022×10²³ molecules / 1 mol
3.0×10²³ molecules
A) Molar mass gold ( Au) = 196.96 g/mol
1 mole Au ----------- 196.96 g
? moles Au ---------- 35.12 g
35.12 x 1 / 196.96
= 0.178 moles of Au
_____________________________
b) 196.96 g --------------- 6.02x10²³ atoms
35.12 g ---------------- ( atoms ? )
35.12 x ( 6.02x10²³) / 196.96
2.114x10²⁵ / 196.96
= 1.07x10²³ atoms
_____________________________________________
Answer:
1.8 g
Explanation:
Step 1: Write the balanced equation
CH₃CH₃(g) + 3.5 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(g)
Step 2: Determine the limiting reactant
The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:112.0 = 0.2684:1.
The experimental mass ratio of CH₃CH₃ to O₂ is 0.60:3.52 = 0.17:1.
Thus, the limiting reactant is CH₃CH₃
Step 3: Calculate the mass of CO₂ produced
The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:88.02.
0.60 g CH₃CH₃ × 88.02 g CO₂/30.06 g CH₃CH₃ = 1.8 g
due to there reactive rate?
