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3 years ago

What is the stable ion of cesium

Chemistry

1 answer:

Sauron [17]3 years ago

5 0

Answer:

Cesium has one valence electron that can be removed from itself resulting to its stable monatomic ion of Cs+ .

Explanation:

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Calculate the pH of a solution prepared by mixing: (Show your work for these calculations) pk of acetic acid is 4.75 a. two mole

kupik [55]

Answer:

Explanation:

To calculate pH you need to use Henderson-Hasselbalch formula:

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

Where HA is the acid concentration and A⁻ is the conjugate base concentration.

The equilibrium of acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75

Where CH₃COOH is the acid and CH₃COO⁻ is the conjugate base.

Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:

pH = 4,75 + log₁₀ $\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}$

a) The pH is:

pH = 4,75 + log₁₀ $\frac{[2 mol]}{[2 mol]}$

pH = 4,75

b) The pH is:

pH = 4,75 + log₁₀ $\frac{[2 mol]}{[1mol]}$

pH = 5,05

I hope it helps!

7 0

4 years ago

An unstable nucleus which starts a decay process is called the _____.

xeze [42]

It Is Called The Parent Nuclide

8 0

4 years ago

Read 2 more answers

Air at 7S°F and14.6 though a cfm(cubic ft per minute) and the The flow rate is 48000 psia flows though a rectangular duct of Ix2

IgorLugansk [536]

Explanation:

The given data is as follows.

Air is at $75^{o}F$ and 14.6 psia.

$\varepsilon$ = 0.00015 ft, Flow rate, (Q) = 48000 $ft^{3}/m$

(a) Formula to calculate hydraulic radius $(r_{H})$ is as follows.

$r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}$

= $\frac{2 \times 1}{2(1) + 2(2)}$

= $\frac{1}{3}$ ft

Formula for equivalent diameter is as follows.

$D_{eq} = 4 \times r_{H}$

= $4 \times \frac{1}{3} ft$

= $\frac{4}{3}$ ft

(b) Formula for velocity floe is as follows.

Q = VA

V = $\frac{Q}{A}$

= $\frac{48000}{2 \times 1}$ ft/min

= 24000 ft/min

(c) Formula to calculate Reynold's number is as follows.

$R_{e}$ = $\frac{D \times V \times \rho}{\mu}$

= $\frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}$ (as $\rho = 0.0744 lb/ft^{3}$ and $\mu$ = 0.0443 lb/ft. hr)

= 53742.66 hr/min

As 1 hr = 10 min. So, $53742.66 hr/min \times \frac{60 min}{1 hr}$

= 3224559.6

(d) Formula to calculate pressure drop $(\Delta P)$ is as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

Putting the given values into the above formula as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

= $\frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}$

= 6.238 $lb/ft^{2}$

5 0

3 years ago

The enthalpy of solution (dissolving) of sodium hydroxide is given below. Determine the change in temperature of a coffee cup ca

geniusboy [140]

Answer:

The change in temperature of a coffee cup calorimeter is 8.87°C.

Explanation:

Volume of the water = V = 150 g

Density of the water , d =1.0 g/mL

Mass of the water = M

$M=d\times V=1.00 g/mL\times 150 ml =150.0 g$

Mass of solution = m = M = 150.0 g

$NaOH(s)\rightarrow NaOH(aq),\Delta H =-44.51 kJ/mol$

Moles of NaOH = $\frac{5.00 g}{40 g/mol}=0.125 mol$

Energy released when 0.125 moles of NaOH added in water = Q

$Q=0.125 moles\times (-44.51 kJ/mol)=-5.5638 kJ=-5,563.8 J$

1 kJ = 1000 J

Heat gained by water = Q' = -Q ( conservation of energy)

$Q'= 5,563.8 J$

Specific heat of solution = c = 4.184 J/g°C

Change in temperature of the solution = $\Delta T$

$Q'=mc\times \Delta T$

$5,563.8 J=150.0 g\times 4.184 J/g^oC\times \Delta T$

$\Delta T=\frac{5,563.8 J}{150.0 g\times 4.184 J/g^oC}=8.87^oC$

The change in temperature of a coffee cup calorimeter is 8.87°C.

7 0

3 years ago

Which statement best describes the structure of covalent network solids? A. They are giant structures held together by weak forc

konstantin123 [22]

Answer: Option (B) is the correct answer.

Explanation:

A covalent compound is a compound formed by sharing of electrons. And, in a covalent network solid atoms are bonded by covalent bonds in a continuous network that is extending throughout the material or solid.

This continuous arrangement of atoms are like a lattice.

For example, diamond is a covalent network solid in which carbon atoms are arranged in a continuous lattice like structure.

Hence, we can conclude that the statement all the atoms are covalently bonded to other atoms to form a lattice-like structure, best describes the structure of covalent network solids.

6 0

3 years ago