Answer:
8.72 × 10^5 moles
Explanation:
To find the number of moles in 5.25 x 10^29 molecules of sucrose, we divide the number of molecules by Avagadro constant (6.02 × 10²³ molecules). That is;
no. of moles = no. of molecules ÷ 6.02 × 10²³ molecules
In this case of sucrose, no of moles contained is as follows;
5.25 × 10^29 ÷ 6.02 × 10²³
5.25/6.02 × 10^ (29-23)
0.872 × 10^6
= 8.72 × 10^5 moles
Answer:
Option A.
2Na + 2H2O —> 2NaOH + H2
Explanation:
To know which option is correct, we shall do a head count of the number of atoms present on both side to see which of them is balanced. This is illustrated below below:
For Option A:
2Na + 2H2O —> 2NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 2 Na
4 H >>>>>>>>>>>> 4 H
2 O >>>>>>>>>>>> 2 O
Thus, the above equation is balanced.
For Option B:
2Na + 2H2O —> NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 1 Na
4 H >>>>>>>>>>>> 3 H
2 O >>>>>>>>>>>> 1 O
Thus, the above equation is not balanced.
For Option C:
2Na + H2O —> 2NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 2 Na
2 H >>>>>>>>>>>> 4 H
1 O >>>>>>>>>>>> 2 O
Thus, the above equation is not balanced.
For Option D:
Na + 2H2O —> NaOH + 2H2
Reactant >>>>>>> Product
1 Na >>>>>>>>>>> 1 Na
4 H >>>>>>>>>>>> 5 H
2 O >>>>>>>>>>>> 1 O
Thus, the above equation is not balanced.
From the illustrations made above, only option A is balanced.
Molar mass of N = 14 g/molMolar mass of O2 = 32 g/molAdding both masses = 46 g/molActual molar mass/ Empirical molar mass = 138.02 / 46 = 3Now multiplying this co effecient with empirical fomula NO2 = 3(NO2) = N3O6So according to above explanation,D) N3O6, is the correct answer.
The answer is 8 :)
All nobel gases have 8 outer electrons.
Answer:
[K₂CrO₄] → 8.1×10⁻⁵ M
Explanation:
First of all, you may know that if you dilute, molarity must decrease.
In the first solution we need to calculate the mmoles:
M = mmol/mL
mL . M = mmol
0.0027 mmol/mL . 3mL = 0.0081 mmoles
These mmoles of potassium chromate are in 3 mL but, it stays in 100 mL too.
New molarity is:
0.0081 mmoles / 100mL = 8.1×10⁻⁵ M
