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AleksAgata [21]
3 years ago
15

Identify the element that matches this orbital diagram

Chemistry
1 answer:
Zolol [24]3 years ago
8 0

Answer:

Nitrogen

Explanation:

The electron configuration of nitrogen is:

1s2 2s2 2p3

which satisfies the diagram.

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Government should use informed science to help make policies protect all citizens<br> True or False
Sloan [31]

Answer:

yes

Explanation:

8 0
3 years ago
WILL GIVE THANKS,STARTS AND BRANLIEST!!!!! PLEASE COME LOOK! SHARE YOUR SMARTNESS!!!! PLSSSSS AT LEAST COME TAKE A LOOK! ANSWER
madreJ [45]

Answer:

B) The group the atom is located

C) There needs to be 2 Na atoms and 1O atom to form Na2O

6 0
3 years ago
Read 2 more answers
A 125G sample of water was heated to 100.0°C and then I borrow platinum 20.0°C is dropped into the beaker the temperature of the
Sophie [7]

Answer:

mass of platinum = 2526.12 g

Explanation:

Given data:

Mass of water = 125 g

Initial temperature of water= 100.0°C

Initial temperature of Pt = 20.0°C

Final temperature = 235°C

Specific heat of Pt = 0.13 j/g°C

Specific heat of water = 4.184 j/g°C

Mass of platinum = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

Q(w) = Q(Pt)

m.c.  (T2 - T1)   =   m.c.   (T2 - T1)

125 g × 4.184 j/g°C ×  (235°C - 100.0°C)  = m × 0.13 j/g°C ×  (235°C - 20°C)

125 g × 4.184 j/g°C × 135°C  = m × 0.13 j/g°C × 215°C

70605 j = m×27.95 j/g

m = 70605 j /27.95 j/g

m = 2526.12 g

5 0
4 years ago
Calculate the pressure kPa of 0.421 mole of helium gas at 254 degree celsius when it occupies a volume of 3.32 L.
Ronch [10]
I didn’t understand the question ♥️
3 0
3 years ago
What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i
UNO [17]

Answer:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)

The net equation is then:

Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-

Actual cathode half-equation:

Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)

Actual net reaction:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

6 0
3 years ago
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