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.
Sound can be used to alert us in emergencies. 2. Sound can also be used to help blind people navigate the world better. I’m sorry if these horrible but I tried lol
Answer: -
6
Explanation: -
The given unbalanced chemical equation is As + NaOH -- > Na3AsO3 + H2
We see there 3 sodium on the right side from Na3AsO3.
But there are only 1 sodium on the left from NaOH.
So we multiply NaOH by 3.
As + 3 NaOH -- > Na3AsO3 + H2
Now we see the number of Hydrogen on the left is 3.
But the number of hydrogens is 2 on the left.
So, we multiply to get both sides 6 hydrogen.
As + 6NaOH -- > Na3AsO3 + 3 H2
Rebalancing for Na,
As + 6NaOH -- > 2Na3AsO3 + 3 H2.
Finally balancing As,
2 As + 6 NaOH -- > 2Na3AsO3 + 3H2
The coefficient of the NaOH molecule in the balanced reaction is thus 6
Answer:
4,38%
small molecular volumes
Decrease
Explanation:
The percent difference between the ideal and real gas is:
(47,8atm - 45,7 atm) / 47,8 atm × 100 = 4,39% ≈ <em>4,38%</em>
This difference is considered significant, and is best explained because argon atoms have relatively <em>small molecular volumes. </em>That produce an increasing in intermolecular forces deviating the system of ideal gas behavior.
Therefore, an increasing in volume will produce an ideal gas behavior. Thus:
If the volume of the container were increased to 2.00 L, you would expect the percent difference between the ideal and real gas to <em>decrease</em>
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I hope it helps!
