<span>Nitrate only. Niritate is anion.</span>
The answer is caso4 and ca(C2H3O2)2
Answer:
3.75 moles
Explanation:
The chemical equation is 2CH₃OH + 3O₂ -> 2CO₂ + 4H₂O
2 moles of CH₃OH are burned by 3 moles of O₂
For 2.5 moles of CH₃OH are burned by x moles of O₂
Let's solve for x :
2*x=2.5*3 => 2*x=7.5 => x=3.75 moles of O₂ are needed to burn 2.5 moles of CH₃OH
We are asked in the problem to convert teh unit mci or millicurie to another unit of radioactivity, becquerel. The conversion to be used in this case is <span>1 millicurie [mCi] = 37,000,000 becquerel [Bq] .The problem gives 15 mCi which when multiplied to </span>37,000,000 is equal to 5,550,000,000 bQ
Answer:
d =~ 5.8μm
d =~ 0.13 μm
Explanation:
when the doping concentrations are 5 × 10^15 cm^-3
d = v^-1/3 ; where d represent the distance between the atoms , and v represent the volume
d =1/ ∛v
d = 1/ ∛5 × 10^15
d = 1/ 170997.5
d = 5.85 × 10 ^ -6
d =~ 5.8μm
when the doping concentrations are 5 × 10^20 cm^-3
d = v^-1/3 ; where d represent the distance between the atoms , and v represent the volume
d =1/ ∛v
d = 1/ ∛5 × 10^20
using the principle of surds and standard forms, we have
d = 1/ ∛0.5 × 10^21
d = 1/7937005.26
d = 1.26 × 10 ^ -7
d = 0.126 × 10 ^ -6
d =~ 0.13 μm