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LenaWriter [7]
3 years ago
6

How many grams of N are in 89.0 g of NH4NO3

Chemistry
1 answer:
yarga [219]3 years ago
8 0
Step 1;calculate  the  molar  mass  of   NH4N03 
molar  mass  of different  element
N=14
H=1
O=16
therefore  molar  mass  of  NH4NO3   is 14+4+14+48= 80g/mol
step 2:find the  molar mass of N  present  in  NH4NO3
   That  is  14+14 =28g/mol
the mass  is  therefore
{(28g/mol/80g/mol) x 89g} =31.15g
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Afina-wow [57]

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3 0
3 years ago
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
Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)
Fynjy0 [20]

Answer:

Rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

Explanation:

According to equation   2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

                     ⇒ -\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}  -----------------------------(1)

    Given that the rate of disappearance of oxygen = -\frac{d[O_{2} ]}{dt} = 3.64 x 10⁻³ M/s

             So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = ?

from equation (1) we can write

                                   \frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]

                                ⇒ \frac{d[SO_{3}] }{dt} = 2 x 3.64 x 10⁻³ M/s

                                ⇒ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

7 0
3 years ago
Using your knowledge of colligative properties explain whether sodium chloride or calcium chloride would be a more effective sub
Lisa [10]
Sodium chloride because it contains the most reactive metal(sodium) and most reactive non-metal(chlorine).
6 0
3 years ago
Mass box A = 10 grams; Mass box B = 5 grams; Mass box C—made of one A and one B
eimsori [14]
2 boxes of A
Because C = A + B
2 of A = 20 grams
at the other hand we have 2 of B = 10
So 20 + 10 = 30 grams
3 0
3 years ago
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