Answer:
hjuijhbhjijnjnjghbgkjfgvv
Explanation:
Answer:
I = 1.23 A
Explanation:
Given that,
The resistance of the lightbulb, R = 96.8 Ω
Voltage, V = 120 V
We need to find the current flows through the lightbulb. Let the current be I. We can use the ohm's law to find it i.e.
So, the current flows through the bulb is 1.23 A.
10 gm of Fe will consumes 19 gm Cl₂ and will produces 29 gm FeCl₃.
What ois Theoretical yield ?
The quantity of a product obtained from a reaction is expressed in terms of the yield of the reaction.
The amount of product predicted by stoichiometry is called the theoretical yield, whereas the amount obtained actually is called the actual yield.
- As 2 moles (111.68 g) of Fe consumes 213 gm of Cl₂ to produce 2FeCl₃
Therefore ,
10 gm of Fe will consumes = 213 / 111.68 x 10 = 19 gm Cl₂
- As 2 moles (111.68 g) of Fe produces 2 mole (324 gm) of FeCl₃
Therefore ,
10 gm of Fe will produces = 324 / 111.68 x 10 = 29 gm FeCl₃
Hence , 10 gm of Fe will consumes 19 gm Cl₂ and will produces 29 gm FeCl₃.
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<u>Answer:</u> The chemical formula of Iron (III) sulfate is 
<u>Explanation:</u>
Iron is the 26th element of periodic table having electronic configuration of ![[Ar]3d^64s^2](https://tex.z-dn.net/?f=%5BAr%5D3d%5E64s%5E2)
.
To form 
ion, this element will loose 3 electrons.
Sulfate ion is a polyatomic ion having chemical formula of 
By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.
So, the chemical formula for Iron (III) sulfate is
Answer:
1) Se2O5
2) I2O6
3)Zn3n2
4) Cr(HCO3)3
Explanation:
selenium pentaoxide (= also called diselenium pentoxide)
= Se2O5
⇒ Se = 78.97 g/mol
⇒ O = 16 g/mol
⇒ 2*78.97 + 5*16 = 237.94 g/mol
iodine trichloride
= I2O6
⇒ I = 126.9 g/mol
⇒ Cl = 35.45 g/mol
⇒ 2* 126.9 + 6 * 35.45 = 466.5 g/mol
zinc (1) nitride does not exist (it's Zinc(ii)nitride
The oxidation number for zinc is always 2
Zn3n2
⇒ Zn = 65.38 g/mol
⇒ N = 14 g/mol
⇒3*65.38 + 2* 14 = 224.14 g/mol
chromium (III) bicarbonate
Cr(HCO3)3
⇒ Cr = 52 g/mol
⇒ H = 1.01 g/mol
⇒ C = 12 g/mol
⇒ O = 16 g/mol
52 + 3*1.01 + 3*12 + 6*16 = 235.03 g/mol
