Answer:
The time required for the coating is 105 s
Explanation:
Zinc undergoes reduction reaction and absorbs two (2) electron ions.
The expression for the mass change at electrode 
is given as :
where;
M = molar mass
Z = ions charge at electrodes
F = Faraday's constant
I = current
A = area
t = time
also; = 
; replacing that into above equation; we have:
---- equation (1)
where;
A = area
d = thickness
= density
From the above equation (1); The time required for coating can be calculated as;
![[ \frac{20 cm^2 *0.0025 cm*7.13g/cm^3}{65.38g/mol}*2 \frac{moles\ of \ electrons}{mole \ of \ Zn} * 9.65*10^4 \frac{C}{mole \ of \ electrons } ] = (20 A) t](https://tex.z-dn.net/?f=%5B%20%5Cfrac%7B20%20cm%5E2%20%2A0.0025%20cm%2A7.13g%2Fcm%5E3%7D%7B65.38g%2Fmol%7D%2A2%20%5Cfrac%7Bmoles%5C%20of%20%5C%20electrons%7D%7Bmole%20%5C%20of%20%5C%20Zn%7D%20%2A%209.65%2A10%5E4%20%5Cfrac%7BC%7D%7Bmole%20%5C%20of%20%5C%20electrons%20%7D%20%20%5D%20%3D%20%2820%20A%29%20t)
= 105 s
Answer:
We have to add 2.30 L of oxygen gas
Explanation:
Step 1: Data given
Initial volume = 4.00 L
Number of moles oxygen gas= 0.864 moles
Temperature = constant
Number of moles of oxygen gas increased to 1.36 moles
Step 2: Calculate new volume
V1/n1 = V2/n2
⇒V1 = the initial volume of the vessel = 4.00 L
⇒n1 = the initial number of moles oxygen gas = 0.864 moles
⇒V2 = the nex volume of the vessel
⇒n2 = the increased number of moles oxygen gas = 1.36 moles
4.00L / 0.864 moles = V2 / 1.36 moles
V2 = 6.30 L
The new volume is 6.30 L
Step 3: Calculate the amount of oxygen gas we have to add
6.30 - 4.00 = 2.30 L
We have to add 2.30 L of oxygen gas
Assuming we have 100 g of sample
30.45/MW of N 14g = 2.175
69.55/MW of O 16g = 4.34
4.34/2.185 = 2
for every 1 mole of N we have 2 moles of O
so the empirical formula would be NO2
without having the molecular weight of the entire molecule the molecular formula can not be determined with the information in your question
Explanation:
a. 0.0093
Number of significant figures = 2
All zero’s preceding the first integers are never significant
b. 120.9
Number of significant figures = 4
All zero’s between integers are always significant.
c. 1,000
Number of significant figures = 1
All zeroes used solely for spacing the decimal point are not significant.
d. 1.008
Number of significant figures = 4
All zero’s between integers are always significant.
All zero’s after the decimal point are always significant.
e. 670
Number of significant figures = 2
All zeroes used solely for spacing the decimal point are not significant.
f. 0.184
Number of significant figures = 3
All zero’s after the decimal point are always significant.
g. 1.30
Number of significant figures = 3
All zero’s after the decimal point are always significant.
Answer:
156 Hydrogen atoms
Explanation:
<u>Any acyclic alkane has a molecular formula that can be expressed as</u>:
CₙH₂ₙ₊₂
Where <em>n</em> is any integer and the number of carbon atoms. For example, Propane has 3 carbon atoms, this means it would have [2*3+2] 8 hydrogen atoms, resulting with a formula of C₃H₈.
An acyclic alkane with 77 carbon atoms would thus have:
2*77 + 2 = 156 hydrogen atoms
