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Lera25 [3.4K]
3 years ago
9

What is the pOH of a 5.6 x 10-5 M solution of cesium hydroxide (CsOH)?

Chemistry
1 answer:
Rudiy273 years ago
7 0

Answer: 4.25

Explanation: CsOH=Cs+

pOH= -log10

= -log10(5.6*10^-5)

= -log10(5.6*10^-5)=-(-4.25)

=4.25

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With regards to Lewis-electron diagrams, which of the following compounds has resonance forms?
andre [41]

Explanation:

The resonance compounds are compounds that have the same position of the atoms (same quantity and elements) but the position of the electrons arround them is different in each resonance compound.

In reality, the compound switches between all the resonance structures it has.

One example of resonance Lewis structures is the Ozone's as can be seen in the figure.

3 0
3 years ago
A cylinder with a moveable piston contains 0.552 mol of gas and has a volume of 259 mL . Part A What will its volume be if an ad
Naddik [55]

Answer:

The new volume will be 367mL

Explanation:

Using PV = nRT

V1 = 259mL = 0.000259L

n1 = 0.552moles

At constant temperature and pressure, the value is

P * 0.000259 = 0.552 * RT ------equation 1

= 0.552 / 0.000259

= 2131.274

V2 = ?

n2 = 0.552 + 0.232

n2 = 0.784mole

Using ideal gas equation,

PV = nRT

P * V2 = 0.784 * RT ---------- equation 2

Combining equations 1 and 2 we have;

V2 = 0.784 / 2131.274

V2 = 0.000367L

V2 = 367mL

7 0
3 years ago
1.
xenn [34]
The answer is OH.

Hope this helps!
4 0
3 years ago
What is the mass of an original 5.60-gram sample of iron-53 that remains unchanged after 25.53 minutes?
olga_2 [115]
Answer is 0.7 g

Reason: Given: mass of Fe = 5.60 gWe know that, half life of Fe-53 = t1/2 (Fe-53) = 8.51 min.
Thus, after 25.53 mins, number of half life passed = 25.53/8.51= 3

Now, Amount left after first half life = 5.60g/2 = 2.80 g
Amount left after second half life = 2.80g/2 = 1.40 g
Amount left after third half-life = 1.40g/2 = 0.7g.
4 0
3 years ago
.A gas occupies 25,3 mL at a pressure of 152 kPa. Find the volume if the pressure is
Strike441 [17]

Answer:

47.36mL

Explanation:

Using Boyles law equation, which states that:

P1V1 = P2V2

Where;

V1 = initial volume (mL)

V2 = final volume (mL)

P1 = initial pressure (atm)

P2 = final pressure (atm)

Based on the provided information, V1 = 25.3mL, P1 = 152 kPa, V2 = ?, P2 = 0.804atm

First, we need to convert 152kPa to atm by dividing by 101

1kPa = 0.0099atm

152kPa = 1.505atm

P1V1 = P2V2

1.505 × 25.3 = 0.804 × V2

38.08 = 0.804V2

V2 = 38.08/0.804

V2 = 47.36mL

5 0
3 years ago
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