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Sever21 [200]
4 years ago
15

What does the electron cloud model describe

Chemistry
1 answer:
erastova [34]4 years ago
7 0
It is used to describe where electrons are when they go around the nucleus of an atom.

C
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A flash contains 85.6g of C7H5NO3S (Saccharine) in 2.00 litre of a solution. What is the molarity?
larisa86 [58]

Answer:

0.234 M

Explanation:

C- 12.009 x 7

H- 1.001 x 5

N- 14.006

O- 16 x 3

S- 32.059

___________+

183.133 g/mol

\frac{1mol}{183.133g} *\frac{85.6g}{2.00L}  = 0.234 M Cancel out the grams mol/L equals molarity. Lowest significant figure is 3

7 0
3 years ago
ICMP scanning involves in checking for the live systems, which can be done by sending the following ping scan request to a host.
abruzzese [7]

Answer:

ICMP Echo Request

Explanation:

ICMP Echo Request is a form of probe or message sent by a user to a destination system.

5 0
4 years ago
Which of the following is/are part of the nitrogen cycle?
Pachacha [2.7K]
Denitrification is part of the nitrogen cycle 
7 0
4 years ago
Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the r
Kazeer [188]

Answer:

∆H° rxn = - 93 kJ

Explanation:

Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds  broken minus bonds formed (H according to Hess Law.

We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.

              N₂ (g)   +            3H₂ (g)   ⇒                          2NH₃ (g)

1 N≡N = 1(945 kJ/mol)     3 H-H = 3 (432 kJ/mol)       6 N-H = 6 ( 389 kJ/mol)

∆H° rxn = ∑  H bonds broken  - ∑ H bonds formed

∆H° rxn = [ 1(945 kJ)   + 3 (432 kJ) ] - [ 6 (389 k J]

∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ

be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond,  N≡N,  we have to use this one .

8 0
3 years ago
Volume occupied 3.52x10^32 moluchles<br>of Mathane (CH4)<br>1) At STP​
Naddika [18.5K]

Answer:

volume = 13097674418.528dm³

Explanation:

n = (3.52)*10^32/(6.02)*10^23)

n = (584717607.97)

n = volume /molar volume

molar volume at stp = 22.4dm³

volume= 584717607.97 x 22.4

volume = 13097674418.528dm³

6 0
3 years ago
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