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Tju [1.3M]
2 years ago
7

Fe + O2 + H₂O- & Fe2O3 + H₂O. balance por el método de tanteo​

Chemistry
1 answer:
AveGali [126]2 years ago
5 0

Answer:

Fe + O2 + H₂O___ Fe2O3 + H₂O.

6Fe+302+H2O______2fe 203+H2O

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Plsz do solv this question hpl​
emmasim [6.3K]
We are given the resistance and voltage of this lamp and we are asked to find the current; the equation that relates these together is Ohm’s Law, V = IR. Simply plug and solve:

V = IR
(220 V) = I(484 Ohms)
I = 0.4545 Amps

The lamp has a current of 0.4545 Amps passing through it under these conditions.

Hope this helps!
7 0
2 years ago
The rate for a reaction between reactants l, m, and n is proportional to the cube of [l] and the square of [m]. what is the rate
postnew [5]
Answer
rate =k[l]^3[m]^2
4 0
3 years ago
17 p<br> 18 n<br> What is the atomic number of this element?
Verdich [7]

Answer:

Chlorine

Explanation:

The P usually stands for Protons, and that is usually the element number. (chlorine is 17) N is for neutron

6 0
3 years ago
What neutral element has the electron configuration described by the Aufbau diagram above.​
devlian [24]

Answer:

Aluminium.

Explanation:

The above electronic configuration can be written in a simplified form as shown below:

1s² 2s²2p⁶ 3s²3p¹

Next, we shall determine the number of electrons in the atom of the element as follow:

Number electron = 2 + 2 + 6 + 2 + 1

Number of electron = 13

Next, we shall determine the number of protons.

Since the element is in its neutral state,

The number of electrons and protons are equal i.e

Proton = Electron

Number of electron = 13

Proton = Electron = 13

Proton = 13

Next, we shall determine the atomic number of the element.

The atomic number of an element is simply the number of protons in the atom of the element i.e

Atomic number = proton number

Proton = 13

Atomic number = 13

Comparing the atomic number of the element with those in the periodic table, the element with the above electronic configuration is aluminium since no two elements have the same atomic number.

5 0
3 years ago
The half-life of cesium-137 is 30 years. suppose we have a 200-mg sample.
Assoli18 [71]
a) To find  the mass after t years:

we will use this formula:

A = Ao / 2^n 

when A =the amount remaining

and Ao = the initial amount

and n = t / t(1/2)

by substitution:

∴ A = 200 mg/ 2^(t/30y)


b) Mass after 90 y :

by  using the previous formula and substitute t by 90 y

A = 200mg/ 2^(90y/30y)

∴ A = 25 mg

C) Time for 1 mg remaining:

when A= Ao/ 2^(t/t(1/2)

so, by substitution:

1 mg = 200 mg / 2^(t/30y)

∴2^(t/30y) = 200 mg  by solving for t

∴ t = 229 y 


7 0
3 years ago
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