Answer: The work done to lift the 2 N rock was 6 Joules.
Explanation:
Weight of the Rock = 2 Newtons
So, the force required to lift this rock will be equal to its weight or greater than its weight
Minimum force required to lift the Rock = 2 N
Displacement of the Rock when force was applied = 3 m,
Work Done: It is defined as product of force applied to the object into displacement of the object.
The work done to lift the 2 N rock was 6 Joules.
The main formulas are <span>pH=pKa + log(Base/Acid) and pKa = </span><span>-log(Ka)
so firstly, we must find the value of pKa,
Ka=6.37 x 10 ^-5, and then logKa= log (6.37 . 10^-5)= -9.66, so -logKa= +9.66=Ka
next let's find </span><span>log(Base/Acid)
for that the concentration of NaOH is [NaOH] = </span><span>500.0 x 0.110 / 500+535 =0.053M, the concentration of the Acid is [Acid] =535*0.25 / </span><span>500+535 =0.12M, so its difference is </span><span><span>[Acid]-</span>[NaOH] = 0.12-0.053=0.07
so pH=</span><span><span>pKa + log(Base/Acid)= 9.66 + log(0.053 / 0.07)= 9.66-0.36=9.29</span> </span>
so pH=9.29.
Answer:
10%
Explanation:
Given data:
Actual yield = 0.13 moles
Percent yield = ?
Solution:
Balanced chemical equation.
3H₂ + N₂ → 2NH₃
First of all we will calculate the mass of given moles of ammonia.
Number of moles = mass / molar mass
Mass = 0.13 × 17 g/mol
Mass = 2.21 g
2.21 g is experimental yield of ammonia.
Hydrogen is limiting reactant because only two moles of hydrogen present.
we will compare the moles of hydrogen and ammonia from balance chemical equation.
H₂ : NH₃
3 : 2
2 : 2/3×2 = 1.33 moles
Mass of ammonia
Mass = number of moles × molar mass
Mass = 1.33 mol × 17 g/mol
Mass = 22.61 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 2.21 g / 22.61 g
Percent yield =9.8% which is almost 10%
Weight
Everything else stays the same. Weight is determined by how much you are being pulled down by gravity
Tropical Zones
*Tropic of Cancer
*Tropic of Capricorn
I cant see it though
