Answer:

Explanation:
Given:
A solution contains one or more of the following ions such as Ag,
and 
Here the Lithium bromide is added to the solution and no precipitate forms
Solution:
Since with LiBr no precipitation takes place therefore Ag+ is absent
Here on adding
to it precipitation takes place.
Precipitate is as follows,

Thus,
is present
When
is added again precipitation takes place.
Therefore the reaction is as follows,

Therefore,
are present in the solution
Answer: 4.96 moles
Explanation:
C5H12 is the chemical formula for pentane, the fifth member of the alkane family.
Given that,
number of moles of C5H12 = ?
Mass in grams = 357.4 g
Molar mass of C5H12 = ?
To get the molar mass of C5H12, use the atomic mass of carbon = 12g; and Hydrogen = 1g
i.e C5H12 = (12 x 5) + (1 x 12)
= 60g + 12g
= 72g/mol
Now, apply the formula
Number of moles = Mass / molar mass
Number of moles = 357.4g / 72g/mol
= 4.96 moles
Thus, 4.96 moles of C5H12 that are contained in 357.4 g of the compound.
Answer:
M of HI = 5.4 M.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
<em>(XMV) acid = (XMV) base.</em>
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HI = (XMV) Ca(OH)₂.</em>
For HI; X = 1, M = ??? M, V = 25.0 mL.
For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.
<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>
Answer:
Ability to conduct electricity
Melting point