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Levart [38]
1 year ago
3

Predict the shape, state the hybridization of the central atom, and give the ideal bond angle(s) and any expected deviations for

:
(d) BiF₅²⁻
Chemistry
1 answer:
FinnZ [79.3K]1 year ago
4 0

BiF₅²⁻ is state the hybridization of the central atom, and give the ideal bond angle(s).

<h3>What is potassium hydroxide?</h3>

Potassium hydroxide, also known as lye, is a chemical substance with the formula KOH. It is also known as caustic potash and is available in a variety of forms such as pellets, flakes, and powders. It has a variety of chemical, industrial, and manufacturing applications.

Potassium hydroxide, sometimes known as caustic potash, is an inorganic chemical with the formula KOH. KOH is a classic strong base, along with sodium hydroxide. It has numerous industrial and specialty applications, the majority of which take advantage of its caustic nature and acid reactivity.

To learn more about potassium hydroxide visit:

brainly.com/question/21904397

#SPJ4

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Which of the following chemical formulas shows a covalent compound? A.CO2 B.MgO C. NaCl D.HBr
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A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re
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Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of \gamma.

\gamma=\frac{C_p}{C_v}

As, C_p=R+C_v

So, \gamma=\frac{R+C_v}{C_v}

Given :

C_v=20.8J/K\\\\R=8.314J/K

\gamma=\frac{8.314+20.8}{20.8}=1.4

Now we have to calculate the initial volume of gas.

Formula used :

P_1V_1=nRT_1

where,

P_1 = initial pressure of gas = 4.25 atm

V_1 = initial volume of gas = ?

T_1 = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)

V_1=5.80L

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

P_1V_1^{\gamma}=P_2V_2^{\gamma}

where,

P_1 = initial pressure of gas = 4.25 atm

P_2 = final pressure of gas = 2.50 atm

V_1 = initial volume of gas = 5.80 L

V_2 = final volume of gas = ?

\gamma = 1.4

Now put all the given values in above formula, we get:

(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}

V_2=8.47L

Now we have to calculate the final temperature of gas.

Formula used :

P_2V_2=nRT_2

where,

P_2 = final pressure of gas = 2.50 atm

V_2 = final volume of gas = 8.47 L

T_2 = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2

T_2=257.9K\approx 258K

Now we have to calculate the work done.

w=nC_v(T_2-T_1)

where,

w = work done = ?

n = number of moles of gas =1.0 mol

T_1 = initial temperature of gas = 300 K

T_2 = final temperature of gas = 258 K

C_v=20.8J/K

Now put all the given values in above formula, we get:

w=(1.0mol)\times (20.8J/K)\times (258-300)K

w=-873.6J

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

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