<h3>
Answer:</h3>
Molecular equation:
2NH₄Br(aq) + Pb(C₂H₃O₂)₂(aq) →2NH₄C₂H₃O₂(aq) + PbBr₂(s)
<h3>
Explanation:</h3>
- The molecular equation of the reaction between aqueous ammonium bromide and aqueous lead (ii) acetate is given by;
2NH₄Br(aq) + Pb(C₂H₃O₂)₂(aq) →2NH₄C₂H₃O₂(aq) + PbBr₂(s)
- This is an example of a double displacement reaction where two ionic compounds exchange ions with each other to form two new compounds or salts.
- It can also be a precipitation reaction due to the formation of precipitate from two ionic compounds. Lead(ii) bromide is the precipitate formed during the reaction.
- The net ionic equation for the reaction is;
2Br⁻(aq) + Pb²⁺(aq)→ PbBr₂(s)
Answer:
0.933 L
Explanation:
Since the pressure is the same, we use the equation
V = Volume
T = Temperature
Since we are given the temperature in Celsius, we need to convert it to Kelvin by adding 273:
-55.0 + 273 = 218
40.0 + 273 = 313
The gas will occupy a volume of 0.933 L.
(Side note - If the temperature increases, the gas will want to expand, leading to a higher volume.)
Answer:
Amount of HCL = 0.00318 L of 3.18 ml
Explanation:
Given:
HCL = 2.5 M
NaOH = 0.53 M
Amount of NaOH = 15 ml = 0.015 L
Find:
Amount of HCL
Computation:
HCL react with NaOH
HCl + NaOH ⇒ NaCl + H₂O
So,
Number of moles = Molarity × volume
Number of moles of NaOH = 0.53 × 0.015
Number of moles of NaOH = 0.00795 moles
So,
Number of moles of HCl needed = 0.00795 mol
es
So,
Volume = No. of moles / Molarity
Amount of HCL = 0.00795 / 2.5
Amount of HCL = 0.00318 L of 3.18 ml
Answer:
<em>The revised periodic table could account for variations resulting from isotopes.</em>
Explanation:
Dmitri Mendeleev was the first person to form the periodic table.
There are certain draw backs in Mendeleev's periodic table which need to be amended.
One such problem is that Mendeleev's periodic table lacks separate place for isotopes of an element. The isotopes share the same place in his periodic table. Hence, if a revised periodic table was formed in which there are separate places for the isotopes of an element then this problem can be amended.