Let us assume that the ring is a size 7 ring, which has a circumference of 54.3 millimeters. Converting this to centimeters, the circumference of the ring is:
54.3 mm = 5.43 cm
Now, we determine the number of gold atoms that will be present in this:
5.43 / 1 x 10⁻⁹
There will be 5.43 x 10⁹ atoms
We now determine the number of moles this is by:
one mole = 6.02 x 10²³ atoms
Moles = 5.43 x 10⁹ / 6.02 x 10²³
Moles = 9.01 x 10⁻¹⁵ moles
The molar mass of gold is 197 g/mol
The mass is 9.01 x 10⁻¹⁵ * 197
The mass of the strand is 1.76 x 10⁻¹² grams
Answer: 0.0000332mol
Explanation: 1mole of CCl4 contains 6.02x10^23 molecules.
Therefore, X mol of CCl4 will contain 2 x 10^19 molecules i.e
Xmol of CCl4 = 2 x 10^19/ 6.02x10^23 = 0.0000332mol
<h3>Answer:</h3>
Excess Reagent = NBr₃
<h3>Solution:</h3>
The Balance Chemical Equation for the reaction of NBr₃ and NaOH is as follow,
2 NBr₃ + 3 NaOH → N₂ + 3 NaBr + 3 HBrO
Calculating the Limiting Reagent,
According to Balance equation,
2 moles NBr₃ reacts with = 3 moles of NaOH
So,
40 moles of NBr₃ will react with = X moles of NaOH
Solving for X,
X = (40 mol × 3 mol) ÷ 2 mol
X = 60 mol of NaOH
It means 40 moles of NBr₃ requires 60 moles of NaOH, while we are provided with 48 moles of NaOH which is Limited. Therefore, NaOH is the limiting reagent and will control the yield of products. And NBr₃ is in excess as some of it is left due to complete consumption of NaOH.
Answer:
N, O, R
Explanation:
Thanks for answering bro, you saved me
Answer:
You would get 19.969 moles
Explanation:
