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Fudgin [204]
3 years ago
11

Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.110 atm to rise

to 0.150 atm .
Chemistry
1 answer:
USPshnik [31]3 years ago
3 0

Answer:

t = 37.1 s

Explanation:

The equation for the reaction is given as;

                  2 N2O5(g)  --> 4 NO2 + O2

Initial:          0.110                  -             -

change:        -2x                  +4x        +x

Final:          0.110 - 2x           +4x        +x

But final = 0.150atm;

0.110 - 2x    +  4x   +  x = 0.150 atm

3x = 0.150 - 0.110

x = 0.0133 atm

Pressure in reactant side;

0.110 - 2x

0.110 - 2 (0.0133) = 0.0834 atm

The integral rate law expression is given as;

ln ( [A] / [Ao] ) = -kt

k =  rate constant = 7.48*10^-3*s-1

ln (0.0834/0.11) = (7.48*10^-3)  t

upon solving, t = 37.1 s

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Answer:

two electrons

Explanation:

Calcium atoms will lose two electrons in order to achieve the noble gas configuration of argon.

8 0
2 years ago
Scientists often investigate alternative explanations for the data and observations presented, even once a hypothesis has been t
worty [1.4K]

Answer:

Explanation:

its 2

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3 years ago
Why won't a fat–soluble vitamin dissolve in water?
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3 years ago
Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point take
tresset_1 [31]

Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = T_f = 165°C

Depression in freezing point = \Delta T_f=?

\Delta T_f=T- T_f=179^oC-165^oC=14^oC

Depression in freezing point  is also given by formula:

\Delta T_f=i\times K_f\times m

K_f = The freezing point depression constant

m = molality of the sample

i = van't Hoff factor

We have: K_f = 40°C kg/mol

i = 1 (  organic compounds)

\Delta T_f=14^oC

14^oC=1\times 40^oC kg/mol\times m

m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg

The molality of isoborneol in camphor is 0.53 mol/kg.

8 0
3 years ago
Which of the following pairs lists a substance that can neutralize H2SO4 and the salt that would be produced from the reaction?
Bogdan [553]

The second option only.

  • LiOH, Li₂SO₄.
<h3>Explanation</h3>

A base neutralizes an acid when the two reacts to produce water and a salt.

Sulfuric acid H₂SO₄ is the acid here. There are more than one classes of bases that can neutralize H₂SO₄. Among the options, there are:

Metal hydroxides

  • Ca(OH)₂ and
  • LiOH.

Metal hydroxides react with sulfuric acid to produce water and the sulfate salt of the metal.

\text{Ca}(\text{OH})_{\bf 2}+\text{H}_2\text{SO}_4 \to \textbf{Ca}\textbf{SO}_{\bf 4} +{\bf 2}\;\text{H}_2\text{O}.

The formula for calcium sulfate \text{CaSO}_4 in option A is spelled incorrectly. Why? The charge on each calcium \text{Ca}^{2+} is +2. The charge on each sulfate ion {\text{SO}_4}^{2-} is -2. Unlike \text{Li}^{+} ions, it takes only one \text{Ca}^{2+} ion to balance the charge on each {\text{SO}_4}^{2-} ion. As a result, \text{Ca}^{2+} and {\text{SO}_4}^{2-} ions in calcium sulfate exist on a 1:1 ratio.

2\;\text{LiOH} +\text{H}_2\text{SO}_4 \to \text{Li}_2\text{SO}_4 + 2\;\text{H}_2\text{O}.

Ammonia, NH₃

Ammonia NH₃ can also act as a base and neutralize acids. NH₃ exists as NH₄OH in water:

\text{NH}_3 + \text{H}_2\text{O} \to \textbf{NH}_{\bf 4}\text{OH}.

The ion {\text{NH}_4}^{+} acts like a metal cation. Similarly to the metal hydroxides, NH₃ (or NH₄OH) neutralizes H₂SO₄ to produce water and a salt:

2\;\textbf{NH}_{\bf 4}\text{OH}+ \text{H}_2\text{SO}_4 \to (\textbf{NH}_{\bf 4})_2\text{SO}_4+2\;\text{H}_2\text{O}.

The formula of the salt (NH₄)₂SO₄ in the fourth option spelled the ammonium ion incorrectly.

As part of the salt (NH₄)₂SO₄, the ammonium ion NH₄⁺ is one of the products of this reaction and can't neutralize H₂SO₄ any further.

7 0
3 years ago
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