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Pie
3 years ago
7

How many moles of each element are in 1 mole of ammonium nitrate?

Chemistry
1 answer:
liubo4ka [24]3 years ago
3 0

Answer:

<h2><u>Of moles of NH4NO3 = 16/80 i.e. 0.2 moles of Ammonium Nitrate. 1 molecule of NH4NO3 contains 1 atom of N. Hence, 1 mole of NH4NO3 would contain 1 mole of Nitrogen atoms and therefore 0.2 moles of NH4NO3 should contain 0.2 moles of Nitrogen atoms</u>.</h2>

Explanation:

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Answer:

It depends

(plum will spoil more quickly at warm temperatures)

Explanation:

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True or False all non - metal elements are gassed at room temperature.​
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Explanation:

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Using these metal ion/metal standard reduction potentials calculate cell potential for Cu2+(aq) + Cd(s) →Cd2+(aq)+ Cu(s) Cu2+(aq
Katena32 [7]

Solution :

Cd(s) ---------------------->  Cd^{+2} (aq) + 2e^-      ,     E_0  =   0.34 v

Cu^{+2}  (aq)  +   2e^-  ------------> Cu (s)             ,     E_0  =  -0.04 v

----------------------------------------------------------------------------------------------

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The cell potential is defined as the measure of \text{ potential difference }  between the \text{two half cells} of an electrochemical cell.  

4 0
3 years ago
What is another name for producer why are they called this​
just olya [345]

Answer:

Autotrophs

Explanation:

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3 0
3 years ago
How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that
tia_tia [17]

Answer:

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,  

= 2.50 L × 0.800 mol/L

= 2 mol

The given pH of a buffer is 8.53

pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,  

= -log (1.8 ×10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:  

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73= 5.37

[NH₄⁺ = 5.37 × 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,  

mass of NH₄Cl = 10.74 mol × 53.5 g/mol

= 574.59 grams.  

8 0
3 years ago
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