Answer:
78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene
Explanation:
Moles of propene = 
According to reaction, 1 mole of propene gives 1 mole of propane.
Then 0.6538 moles of bromo-propane will give:
78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene.
A trustable source was above to say that the first experiments will decide how many isotopes of this element exist. Isotopes are atoms of an element which bear chemically the identical but have dissimilar physical properties. One adaptable property among isotopes is their atomic mass.
Answer:
The formula of the original halide is SrCl₂.
Explanation:
- The balanced equation of this reaction is:
SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X is the halide.
- From the equation stichiometry, 1.0 mole of strontium halide will result in 1.0 mole of SrSO₄.
- The number of moles of SrSO₄ <em>(n = mass/molar mass) </em>= (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
- The number of moles of SrX are 4.11 x 10⁻³ moles from the stichiometry of the balanced equation.
- n = mass / molar mass, n = 4.11 x 10⁻³ moles and mass = 0.652 g.
- The molar mass of SrX₂ = mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
- The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
- The atomic mass of halide X = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2 g/mole = 35.5 g/mole.
- This is the atomic mass of Cl.
- <em>So, the formula of the original halide is SrCl₂</em>.
Answer: not to be rude but who is Miley partridge? If she is a friend then u r a kind hearted person
Explanation:
Answer:
C. 0.20 M Mg ion & 0.40 M Cl ion
Explanation:
MgCl₂ is a ionic salt which is dissociated as this
MgCl₂ → Mg²⁺ + 2Cl⁻
First of all, we have a solution of 200 mL, with [MgCl₂] = 0.6M
Molarity . volume = moles.
0.6 mol/l . 0.2l = 0.12 mol
MgCl₂ → Mg²⁺ + 2Cl⁻
0.12mol 0.12 0.24
This moles are also in 400mL of water, so the new concentration is
[Mg²⁺] = 0.12 m/0.6L = 0.2M
[Cl⁻] = 0.24 m/0.6L = 0.4M
Remember we initially have 200mL and then, we add 400 mL, so we supose aditive volume. (600mL)
