Answer:
Nickle
Explanation:
In statement it is given that the metal "M" has a formula "MCl₂" which means it has a charge of +2 as each Chlorine atom has -1 charge. Therefore, it is cleared that this metal has lost 2 electrons and originally it was carrying 28 electrons.
After that periodic table is concerned and a metal with 28 atomic number was tracked because a neutral atom with 28 electrons must have 28 protons and protons are infact the atomic number.
Hence, in periodic table the metal with 28 atomic number is found to be Nickle.
Reactions of Ethyl-3-pentenoate with all given reagents are given below.
Reaction with H₂ / Pd:
The non-polar double bond present in Ethyl-3-pentenoate is reduced to saturated chain. This reagent can not reduce the carbonyl group.
Reaction with NaBH₄: Sodium Borohydride is a weak reducing agent at compared to LiAlH₄. It can only reduce aldehydes and Ketones to corresponding alcohols.
Reaction with LiAlH₄: Lithium Aluminium hydride is a strong reducing agent. It can reduce all types of carbonyl compounds to corresponding alcohols, But, it can not reduce non-polar double bonds like alkenes and alkynes.
Result: The correct answer is
Option-A (Highlighted RED below).
Answer: The oxidation state of selenium in SeO3 is +6
Explanation:
SeO3 is the chemical formula for selenium trioxide.
- The oxidation state of SeO3 = 0 (since it is stable and with no charge)
- the oxidation number of oxygen (O) IN SeO3 is -2
- the oxidation state of selenium in SeO3 = Z (let unknown value be Z)
Hence, SeO3 = 0
Z + (-2 x 3) = 0
Z + (-6) = 0
Z - 6 = 0
Z = 0 + 6
Z = +6
Thus, the oxidation state of selenium in SeO3 is +6
Answer:
the change is evaporation
Explanation:
the water heats up at the surface of the water and evaporates
Answer:
58.6 % by mass of Na₂CO₃
Explanation:
This is the reaction:
Na₂CO₃ + MgCO₃ + 4HCl → MgCl₂ + 2NaCl + 2CO₂ + 2H₂O
Let's find out the moles of CO₂ produced, by the Ideal Gases Law
1.24 atm . 1.67 L = n . 0.082 . 299K
(1.24 atm . 1.67 L / 0.082 . 299K) = n
0.0844 moles = n
Ratio is 2:1, so 2 moles of dioxide were produced by 1 mol of sodium carbonate. Let's make a rule of three:
2 moles of CO₂ were produced by 1 mol of Na₂CO₃
Then, 0.0844 moles of Co₂ would beeen produced by (0.0844 .1)/2 = 0.0422 moles of Na₂CO₃.
Let's convert this moles into mass (mol . molar mass)
0.0422 mol . 106 g/mol = 4.47 g
Finally we can know the mass percent of sodium carbonate in the mixture
(Mass of compound /Total mass) . 100 → (4.47 g / 7.63g) . 100 = 58.6 %