The term which is described as a long, narrow depression in the ocean floor would be ocean trench. They <span>are hemispheric-scale long but narrow topographic depressions of the sea floor. They are also the deepest parts of the ocean floor. Hope this answers the question.</span>
The number of molecules in one mole of any substance is equal to Avagadro's number
Avagadro's number is 6.023 X 10²³
Thus
1 mole = 6.023 X 10²³ molecules
for ammonia we are provided with three moles
so to obtain the total number of molecules of ammonia in three moles we will multiply the Avagadro's number with three
total molecules = 3 X 6.023 X 10²³
Total molecules of ammonia = 18.069 X 10²³
In scientific notation
Total molecules of ammonia = 1.8069 X 10²⁴ = 1.81 X 10²⁴
Answer:
Explanation:
Hello,
In this case, we use the ideal gas equation to compute the volume as shown below:
Nonetheless we are given mass, for that reason we must compute the moles of gaseous fluorine (molar mass: 38 g/mol) as shown below:
Thus, we compute the volume with the proper ideal gas constant, R:
Best regards.
Answer is: 3,4
Chemical reaction: HNO₂ + NaOH → NaNO₂ + H₂O.
c₀(HNO₂) = 1,2 M = 1,2 mol/dm³.
c₀(NaNO₂) = 0,8 M = 0,8 mol/dm³.
V₀(HNO₂) = V₀(NaNO₂) = 1 dm³ = 1 L.
c₀(NaOH) = 0,5 M = 0,5 mol/dm³.
n₀(HNO₂)= 1,2 mol/dm³ · 1 dm³ = 1,2 mol.
n₀(NaNO₂) = 0,8 mol/dm³ · 1 dm³ = 0,8 mol.
V(NaOH) = 400 mL · 0,001 dm³/mL = 0,4 dm³.
n₀(NaOH) = c₀(NaOH) · V₀(NaOH).
n₀(NaOH) = 0,5 mol/dm³ · 0,4 dm³ = 0,2 mol.
n(HNO₂) = 1,2 mol - 0,2 mol = 1 mol.
n(NaNO₂) = 0,8 mol + 0,2 mol = 1 mol.
c(HNO₂) = 1 mol ÷ 1,4 dm³ = 0,714 mol/dm³.
c(NaNO₂) = 1 mol ÷ 1,4 dm³ = 0,714 mol/dm³.
pH = pKa + log (c(HNO₂) / c(NaNO₂)).
pH = 3,4 + log (0,714 mol/dm³ / 0,714 mol/dm³) = 3,4.
the answer is 0.000097 KM
