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kap26 [50]
3 years ago
12

Identify the following silver salts as soluble or insoluble in water. A g N O 3 AgNOX3 Choose... A g C l AgCl Choose... A g I Ag

I Choose...
Chemistry
1 answer:
maxonik [38]3 years ago
5 0

Answer:

AgNO3 -soluble in water

AgCl- insoluble in water

AgI- Insoluble in water

Explanation:

The solubility of chemical compounds in water is easily predicted by a set of rules generally referred to as the solubility rules. These rules are usually based on experimental observation of diverse groups of compounds.

According to the solubility rules, nitrates are soluble in water including the nitrates of silver. However, halides are soluble in water except those of silver, lead and mercury.

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What is the pH of a solution of RbOH with a concentration of 0.86 M? Answer to 2 decimal places
lubasha [3.4K]

Answer:The pH of the solution is given by pH=−log([H3O+])

Explanation:so you can't use

pH

=

−

log

(

0.150

)

because that's the concentration of the hydroxide anions,

OH

−

, not of the hydronium cations,

H

3

O

+

. In essence, you calculated the

pOH

of the solution, not its

pH

.

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a

1

:

1

mole ratio.

NaOH

(

a

q

)

→

Na

+

(

a

q

)

+

OH

−

(

a

q

)

So your solution has

[

OH

−

]

=

[

NaOH

]

=

0.150 M

Now, the

pOH

of the solution can be calculated by using

pOH

=

−

log

(

[

OH

−

]

)

−−−−−−−−−−−−−−−−−−−−

In your case, you have

pOH

=

−

log

(

0.150

)

=

0.824

Now, an aqueous solution at

25

∘

C

has

pH + pOH

=

14

−−−−−−−−−−−−−−so you can't use

pH

=

−

log

(

0.150

)

because that's the concentration of the hydroxide anions,

OH

−

, not of the hydronium cations,

H

3

O

+

. In essence, you calculated the

pOH

of the solution, not its

pH

.

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a

1

:

1

mole ratio.

NaOH

(

a

q

)

→

Na

+

(

a

q

)

+

OH

−

(

a

q

)

So your solution has

[

OH

−

]

=

[

NaOH

]

=

0.150 M

Now, the

pOH

of the solution can be calculated by using

pOH

=

−

log

(

[

OH

−

]

)

−−−−−−−−−−−−−−−−−−−−

In your case, you have

pOH

=

−

log

(

0.150

)

=

0.824

Now, an aqueous solution at

25

∘

C

has

pH + pOH

=

14

−−−−−−−−−−−−−−

3 0
2 years ago
Awdawdawdadawdanodawdadwadadaw
crimeas [40]
No, actually adawadawada and awawawaw usually addawadadaw but also awawawa so it’s a possibility but very rare.
3 0
2 years ago
I need help this is 7th grade science
GREYUIT [131]

Answer:

i need to know the question then i can help :)

Explanation:

8 0
2 years ago
A car traveling with constant speed travels 150 km in 7200 s. What is the speed of the car?
Advocard [28]
Speed=distance/time  

D=150km

S=7200s 

Divide the numbers. 

150/7200=0.02km/s <---final answer
5 0
3 years ago
If the initial [NO2] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M. If the initial is 0.260 , it
Nitella [24]

The question is incomplete, here is the complete question:

At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas

NO_2\rightarrow NO+\frac{1}{2}O_2

The reaction is second order for NO_2 with a rate constant of 0.543M^{-1}s^{-1} at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M

a) 1.01    b) 5.19     c) 0.299      d) 0.0880     e) 3.34

<u>Answer:</u> The time taken is 5.19 seconds

<u>Explanation:</u>

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.543M^{-1}s^{-1}

t = time taken  = ?

[A] = concentration of substance after time 't' = 0.150 M

[A]_o = Initial concentration = 0.260 M

Putting values in above equation, we get:

0.543=\frac{1}{t}\left (\frac{1}{(0.150)}-\frac{1}{(0.260)}\right)\\\\t=5.19s

Hence, the time taken is 5.19 seconds

6 0
3 years ago
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