It would repel some of the negatively charged electrons of the wall so the answer would be C. The wall and the ballon repel each other
a. mol O₂=0.5
b. volume O₂ = 25 cm³
c. i. the total volume of the two reactants = 75 cm³
c. ii. the volume of nitrogen dioxide formed = 50 cm³
<h3>Further explanation</h3>
Reaction
2NO(gas) + O₂(gas) ⇒ 2NO₂ (gas)
a.
mol NO = 1
From the equation, mol ratio NO : O₂ = 2 : 1, so mol O₂ :
b.
From Avogadro's hypothesis, at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles
Because mol ratio NO : O₂ = 2 : 1, so volume O₂ :
c.
i. total volume of reactants : 25 cm³+ 50 cm³=75 cm³
ii. the volume of nitrogen dioxide formed :
mol ratio NO : NO₂ = 2 : 2, so volume NO₂ = volume NO = 50 cm³
A. conduction
that is the correct answer
Answer:
18.84 g of silver.
Explanation:
We'll begin by calculating the number atoms present in 5.59 g of sulphur. This can be obtained as follow:
From Avogadro's hypothesis,
1 mole of sulphur contains 6.02×10²³ atoms.
1 mole of sulphur = 32 g
Thus,
32 g of sulphur contains 6.02×10²³ atoms.
Therefore, 5.59 g of sulphur will contain = (5.59 × 6.02×10²³) / 32 = 1.05×10²³ atoms.
From the calculations made above, 5.59 g of sulphur contains 1.05×10²³ atoms.
Finally, we shall determine the mass of silver that contains 1.05×10²³ atoms.
This is illustrated below:
1 mole of silver = 6.02×10²³ atoms.
1 mole of silver = 108 g
108 g of silver contains 6.02×10²³ atoms.
Therefore, Xg of silver will contain 1.05×10²³ atoms i.e
Xg of silver = (108 × 1.05×10²³)/6.02×10²³
Xg of silver = 18.84 g
Thus, 18.84 g of silver contains the same number of atoms (i.e 1.05×10²³ atoms) as 5.59 g of sulfur
Answer:
See explanation
Explanation:
Qualitative analysis in chemistry is a method used to determine the ions present in a solution chiefly by means of chemical reactions.
In this case, I suspect the presence of silver ions and/or barium ions. The first step is to add dilute HCl. This will lead to the precipitation of the silver ion as AgCl. If a white precipitate is formed upon addition of HCl then Ag^+ is present in the solution.
Secondly, I add a carbonate such as NH4CO3(aq). This will cause the barium ions to become precipitated as barium carbonate. Hence, the formation of a white precipitate when NH4CO3(aq) is added to the solution indicates the presence of barium ion in the solution.
