Answer:
1.59mol/L
Explanation:
Data obtained from the question include:
Mass of MgCl2 = 151g
Volume of water(solvent) = 1L
Now, let us calculate the number of mole of MgCl2. This is illustrated below:
Molarity Mass of MgCl2 = 24 + (2x35.5) = 24 + 71 = 95g/mol
Mass of MgCl2 = 151g
Number of mole of MgCl2 =?
Number of mole = Mass /Molar Mass
Number of mole of MgCl2 = 151/95
Number of mole of MgCl2 = 1.59mole
Now we can calculate the molarity of MgCl2 as follow:
Mole = 1.59mole
Volume = 1L
Molarity =?
Molarity = mole /Volume
Molarity = 1.59/1
Molarity = 1.59mol/L
Answer:
Find the list and explanation below.
Explanation:
Hydrochloric acid shortened as HCl is a harmful chemical when not handled properly. Four ways to avoid injury with hydrochloric acid include;
1. Dilution: It is advised to dilute the concentrated hydrochloric acid with a reasonable amount of water so that its corrosive effect can be significantly reduced.
2. Wear a laboratory coat: When working with hydrochloric acid in the laboratory it is always advised to wear a laboratory suit so as to ensure a minimal amount of harm in case of an accident or spillage with the acid.
3. Wear a face shield: It is also important that a face shield is worn while working with the acid to prevent contact with the eyes.
4. Wear a protective rubber glove: The rubber gloves would ensure that there is minimal contact of the acid with the hands.
Three steps to be taken in case of an accidental exposure include;
1. Medical Attention: In serious cases, such as in a case of ingesting the acid, urgent medical help should be sought for immediately.
2. Water: If there is contact with the skin or eye, a lot of water should be used on the affected area, and then medical help obtained.
3. Inhalation of fresh air: In a case whereby the hydrochloric acid was inhaled for a long time, the affected person should try to get some fresh air and if other effects are noticed, urgent medical attention should be gotten.
Pure water does
not have enough ions to conduct electricity. A mixture of metals such as iron,
zinc and copper in the wet soil can trigger electrolysis that requires excess
energy in the form of over potential to conduct electricity. The excess energy
is needed due to limited self-ionization of water. The wet soil then can
conduct current when positive and negative ions are present. The water ions begin
to flow from anode (positive electrode) to cathode (negative electrode) to be oxidize
and produce electricity.
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Ammonium carbonate will form 3 moles of ions.
Methyl alcohol will form 0 moles of ions.
Methane will form 0 moles of ions.
Aluminum sulfite will form 3 moles of ions.
Hydrobromic acid will form 2 moles of ions.
Explanation:
One mole of ammonium carbonate will form 3 moles of ions when dissolved in water.
(NH₄)₂CO₃ (s) + H₂O (l) → 2 NH₄⁺ (aq) + CO₃²⁻ (aq) + H₂O (l)
One mole of methyl alcohol will form 0 moles of ions when dissolved in water.
(actually it form ions because of its acidic behavior but they are at the order of 10⁻⁷ moles, however in the framework of this question we may say that there are none)
One mole of methane will form 0 moles of ions when dissolved in water.
Methane does not react with water (in normal conditions) so will not form ions.
One mole of aluminum sulfite will form 3 moles of ions when dissolved in water.
Al₂SO₃ (s) + H₂O (l) → 2 Al₃⁺ (aq) + SO₃²⁻ (aq) + H₂O (l)
One mole of hydrobromic acid will form 2 moles of ions when dissolved in water.
HBr (l) + H₂O (l) → Br⁻ (aq) + H₃O⁺ (aq)
Learn more about:
solvation of ions
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Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane