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Oksanka [162]
3 years ago
7

Given that 7.25 moles of carbon monoxide gas are present in a container of volume 11.90 L, what is the pressure of the gas (in a

tm) if the temperature is 87°C?
Chemistry
1 answer:
d1i1m1o1n [39]3 years ago
6 0

Answer:17.955atm

Explanation:Pv=nrt

P= nrt/v

P= 7.25*0.08205*360/11.90

P= 214.1505/11.90

P=17.995atm

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The population of wild turkeys in an area tripled over the course of three years. Which of the following could be a contributing
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Decrease in predators because then the wild turkeys won't be killed.
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Using a Laptop computer

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How many grams of bromine are required to react completely with 37.4 grams aluminum chloride?
Gekata [30.6K]

Answer:

None  

Explanation:

Cl₂ is above Br₂ in the activity series.

Bromine will not displace chlorine from its salts.

The reaction will not occur.

8 0
3 years ago
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Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv
Triss [41]

Answer:

1. 3.57\times 10^{-4}was the K_a value calculated by the student.

2. 5.93\times 10^{-4}was the K_b of ethylamine value calculated by the student.

Explanation:

1.

The pH value of Aspirin solution = 2.62

pH=-\log[H^+]

[H^+]=10^{-2.62}=0.00240 M

Moles of s asprin = \frac{2.00 g}{180 g/mol}=0.01111 mol

Volume of the solution = 0.600 L

The initial concentration of Aspirin  = c = \frac{0.01111 mol}{0.600 L}=0.0185 M

HAs\rightleftharpoons As^-+H^+

initially

c       0    0

At equilibrium

(c-x)      x   x

The expression of dissociation constant :

K_a=\frac{[As^-][H^+]}{[HAs]}:

K_a=\frac{x\times x }{(c-x)}

=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}

K_a=3.57\times 10^{-4}

3.57\times 10^{-4}was the K_a value calculated by the student.

2.

The pH value of ethylamine = 11.87

pH+pOH=14

pOH=14-11.87=2.13

pOH=-\log[OH^-]

[OH^-]=10^{-2.13}=0.00741 M

The initial concentration of ethylamine = c = 0.100 M

C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-

initially

c                    0    0

At equilibrium

(c-x)                x   x

The expression of dissociation constant :

K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}:

K_b=\frac{x\times x}{(c-x)}

=\frac{0.00741\times 0.00741}{(0.100-0.00741)}

K_b=5.93\times 10^{-4}

5.93\times 10^{-4}was the K_b of ethylamine value calculated by the student.

3 0
3 years ago
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