Question

. Balance these equations:
(a) ____Fe(s) + ____O2(g)h____ Fe2O3(s)
(b) ____H2(g) + ____N2(g)h____NH3(g)
(c) ____Cl2(g) + ____KBr(aq)h____Br2(l ) +____KCl (aq)
(d) ____CH4(g) + ____O2(g)h____CO2(g) +____H2O(l )
@diolle

Answer 1

what's the formula in finding the amu?

Answer 2

????

Answer 3

can you help me answer this?
The element nitrogen has an atomic weight of 14.0 and consists of two stable isotopes nitrogen-14 and nitrogen-15.

The isotope nitrogen-14 has an atomic mass of 14.0 amu and a percent natural abundance of 99.6 %.
The isotope nitrogen-15 has a percent natural abundance of 0.370 %.

What is the atomic mass of nitrogen-15?________amu

Answer 4

?

Answer 5

14.0

Answer 6

wrong

Answer 7

oh it's 15

Answer 8

yeah

Answer 9

just solved it.

Answer 10

i just solved what?

Answer 11

\[[14.0 - 0.996 14.0]/3.70e=15.0 amu\]
is that right?

Answer 12

ok but hoe dp i balance these?

Answer 13

sorry can't help you with that. I don't know that one

Answer 14

yes that is right

Answer 15

can anyone balance these equations i don't get it?

Answer 16

Alright, so you just have to balance these so that they are equal on both sides:

a) \[4Fe + 3O _{2} \rightarrow 2Fe _{2}O_{3}\]

b) 3H2 + N2 = 2NH3

c) Cl2 + 2KBr = Br2 + 2KCl

d) CH4 + 2O2 = CO2 + 2H20

Answer 17

Like in a), all you have to do is find a common multiple between 3 and 2 (6) to balance your oxygen on both sides of the equation. So if you multiply Fe2O3 by 2, you will have 2FeO3 (2 Fe, and 6 O) Likewise, if you multiply O2 by a coefficent of 3, you will have 6 O on that side, thus balancing the oxygen. Then, simply balance the remaining Fe. There are 4 Fe (2 x 2) on the products side, so you must have a coefficient of 4 on the reactants side.

Tip: It may be easier to balance your compounds first, especially ones containing O, and then tackle the elements and ions.

Hope this helps!