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Alex_Xolod [135]
1 year ago
8

According to the bohr model of the atom, the single electron in what motion of a hydrogen atom circles the nucleus

Chemistry
1 answer:
Lady bird [3.3K]1 year ago
4 0
It would be elliptical according to him.....
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Select all that reasons that the reaction contents does not conduct at this low point.
Alexxx [7]

It's all three of the answers

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1 year ago
El amoniaco y el fluor reaccionan para formar tetrafluoruro de dinitrogeno y fluoruro de hidrogeno. segun la reaccion: NH3 + F2
White raven [17]

Answer:

son 12.6 gramos de HF

Explanation:

Tienes que saber qual es el reactor limitante en este caso es fluoruro con los 20 gramos puedes producer .631 mol qual son 12.6 gramos

5 0
2 years ago
If a particular ore contains 55.4 % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosp
gizmo_the_mogwai [7]
 The ore contains 55.4% calcium phosphate (related to the mineral apatite) so the amount of Ca3(PO4)2 is 55.4%x=1000g so x=1000/0.554= 1.805kg. Now for the % of P in this amount of calcium phosphate, use all the masses of the elements in Ca3PO4= Ca=40.078 x 3= 120.23 and (PO4)2= (30.974+64)2=189.95  (NB oxygen is 16 mass x 4 =64) so the total mass is 310.2 and we have 61.95 of P (Pmass x 2) so 61.95/3102.= 0.19 or 19% P. So of the 1.805 x 0.19= 0.34kg of phosphorus.
8 0
2 years ago
How many grams of ca(no3)2 can be produced by reacting excess hno3 with 6.33 g of ca(oh)2?
xeze [42]

Answer:

Amount of Ca(NO3)2 produced = 14.02 g

Explanation:

The given reaction can be depicted as follows:

Ca(OH)2 + 2HNO3 → Ca(NO3)2 + 2H2O

Since it is given that HNO3 is in excess, the limiting reactant is Ca(OH)2

Now, Mass of Ca(OH)2 = 6.33 g

Molar mass of Ca(OH)2 = 74 g/mol

Moles\ Ca(OH)2 = \frac{Mass}{Molar\ Mass} = \frac{6.33 g}{74 g/mol} =0.0855

Based on the reaction stoichiometry:

1 mole of Ca(OH)2 forms 1 mole of Ca(NO3)2

Therefore, moles of Ca(NO3)2 produced from the moles of Ca(OH)2 reacted = 0.0855 moles

Molar mass of Ca(NO3)2 = 164 g/mol

Mass\ Ca(NO3)2 \ produced = moles*molar\ mass \\= 0.0855\ moles*164\ g/mol = 14.02\  g

6 0
2 years ago
Read 2 more answers
What molarity should the stock solution be if you want to dilute 25.0 ml to 2.00 l and have the final concentration be 0.103 m?
andreyandreev [35.5K]
Pls refer the image for solution i have solved it.

8 0
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