I really need help plz plz plz plz plz plz

State the oxidation number assigned to each bold element in the formula: NH4+1 a 3 b -3 c -1 d 6

Leya [2.2K]

The fomula is NH4 (1+)

There are only two elements N and H.

As per oxidation state rules, the most electronegative element will have a negative oxidation state and the other element will have a positive oxidation state.

N is more electronative than H, so H will have a positive oxidation state and nitrogen will have a negative oxidation state.

You can also use the rule that states the hydrogen mostly has 1+ oxidation state,except when it is bonded to metals.

In conclusion the oxidation state of H in NH4 (1+) is 1+.

Now you must know that the sum of the oxidations states equals the charge of the ion, which in this case is 1+.

That implies that 4* (1+) + x = 1+

=> x = (1+) - 4(+) = 3-

Answer: the oxidation state of N is 3-, that is the option b.

8 0

3 years ago

Here is my question only answer if you understand this question

sasho [114]

1- false
2- true
3- true
4- false
5- true
6- false
7- true
8- true
9- true
10- false

Hope it helps :)

8 0

3 years ago

How many neutrons does the element in group 15 and period 4 have ?

alexira [117]

The answer is 19 ahhahaha

4 0

3 years ago

How do i solve these chemistry problems

Ivahew [28]

What you have to do is balance the chemical equations to make sure everything is even on both sides. If you want me to help you answer the questions comment back

8 0

3 years ago

In the gas-phase reaction 2 A + B ⇌ 3 C + 2 D, it was found that, when 1.00 mol A, 2.00 mol B, and 1.00 mol D were mixed and all

Naddik [55]

Answer:

(i)The mole fractions are :

$A=\frac{0.4}{4.3} \\=0.0930$

$B=\frac{1.4}{4.3} \\=0.3256$

$C=\frac{0.9}{4.3} \\=0.2093$

$D=\frac{1.6}{4.3} \\=0.3721$

(ii) $K=0.4508$

(iii)ΔG = 1.974kJ

Explanation:

The given equation is :

$2A+B$ ⇄ $3C+2D$

Let $\alpha$ be the number of moles dissociated per mole of $B$

Thus ,

The initial number of moles of :

$A=1$

$B=2$

$C=0$

$D=1$

$2A\\(1-2\alpha)$ + $B\\2(1-\alpha)$ ⇄ $3C\\(3\alpha)$ + $2D\\(1+2\alpha)$

And finally the number of moles of $C[tex] is 0.9Thus ,[tex]3\alpha=0.9\\\alpha=0.3[tex]The final number of moles of:[tex]A = 1-2\alpha=1-2*0.3=0.4mol[tex] [tex]B=2(1-\alpha)=2(1-0.3)=1.4mol[tex][tex]D=1+2\alpha=1+2*0.3=1.6mol[tex]Thus , total number of moles are : 0.4+1.4+0.9+1.6=4.3(i)The mole fractions are : [tex]A=\frac{0.4}{4.3} \\=0.0930$