Answer:
<h2>8.5 g/cm³</h2>
Explanation:
The density of a substance can be found by using the formula
From the question we have
We have the final answer as
<h3>8.5 g/cm³</h3>
Hope this helps you
Answer:
1 gram of H2 will be produced from 12 grams of Mg.
Explanation:
According to Stoichiometry, 0.5 moles of Mg are present. 1 mole of Mg produces 1 mole of H2, so 0.5 moles of Mg will produce 0.5 moles of H2. Multiplying molar mass of H2 i.e. 2 gram/mole with 0.5 moles, we can find the mass of H2 in grams which is 1 gram.
The question is improperly formatted.
What is the concentration of H+ ions in a 2.2 M solution of HNO3.
Answer:-
2.2 moles of H+ per litre
Explanation:-
M stands for molarity. 2.2 M means 2.2 moles of HNO3 is present per litre of the solution.
Now HNO3 has just 1 H in it's formula. HNO3 would give H+. So 2.2 moles of HNO3 would mean 2.2 moles of H+ per litre.
PH of a solution will be <span>higher than 7
</span>
Ammonium cyanide is a salt formed by hydrogen cyanide and ammonia. Ammonia is a weak base and hydrogen cyanide is a weak acid.
NH₄CN + H₂O ⇒ NH₃ + HCN
NH₄⁺ + H₂O -----> H₃O⁺ + NH₃
CN⁻ + H₂O -----> HCN + OH⁻
Although both compounds are weak electrolytes, NH₃ is somewhat stronger base than HCN is a strong acid, so the solution reacts alkaline. We can prove this using Ka and Kb values:
Ka(HCN) = 4.9 x × 10⁻¹⁰
Kb(NH₃) = 1.8 × 10⁻⁵<span>
Kw= </span>1.0 × 10⁻¹⁴
Let's first calculate Ka for NH₄⁺:
Ka(NH₄⁺) x Kb(NH₃<span>) = pKw
</span>Ka(NH₄⁺) = Kw/Kb(NH₃) = 5.6 x 10⁻¹⁰
Then, Kb for CN⁻:
Kb(CN⁻) x Ka(HCN) = pKw
Kb(CN⁻) = Kw/Ka(HCN) = 2 x 10⁻⁵
From this, we can see that the acid constant NH4⁺ is much lower than the base constant of CN⁻, which will say that the solution of NH₄CN will react slightly alkaline because of the higher presence of hydroxyl ions in solution.
Answer:
A. There was still 140 ml of volume available for the reaction
Explanation:
According to Avogadro's law, we have that equal volumes of all gases contains equal number of molecules
According to the ideal gas law, we have;
The pressure exerted by a gas, P = n·R·T/V
Where;
n = The number of moles
T = The temperature of the gas
R = The universal gas constant
V = The volume of the gas
Therefore, given that the volumes and number of moles of the removed air and added HCl are the same, the pressure and therefore, the volume available for the reaction will remain the same
There will still be the same volume available for the reaction.
