Answer:
1.38×10^25 molecules
Explanation:
Applying n= (no. of molcules)/NA
23 = N/6.02×10^23
= 1.38×10^25 molecules
It would be the top soil. The most dissimilar layer is the uppermost layer.
Answer:
The final balanced equation is
Ni2+ + 2NaOH --> Ni (OH)2 + 2Na+
Explanation:
It is given that sodium hydroxide is added to collect the solid nickel(II) hydroxide product
The empirical equation for this statement is
Ni2+ + NaOH --> Ni (OH)2 + Na+
We will first balance the hydroxide molecule. On the right side there are two OH molecules.
Thus, on the left side we will take 2 sodium hydroxide
Ni2+ + 2NaOH --> Ni (OH)2 + Na+
Now we will balance the sodium ion which are 2 in numbers on the left side and 1 on the right side
Ni2+ + 2NaOH --> Ni (OH)2 + 2Na+
So, the final balanced equation is
Ni2+ + 2NaOH --> Ni (OH)2 + 2Na+
Answer:
Option A. 107 mL
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 150 mL
Initial pressure (P₁) = 500 mmHg
Final pressure (P₂) = 700 mmHg
Temperature = constant
Final volume (V₂) =?
The final volume of the gas can be obtained by using the Boyle's law equation as shown below:
P₁V₁ = P₂V₂
500 × 150 = 700 × V₂
75000 = 700 × V₂
Divide both side by 700
V₂ = 75000 / 700
V₂ = 107 mL
Therefore, the final volume of the gas is 107 mL.
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
