Balance the following equation (separate your answers with commas):

What atomic or hybrid orbital on sb makes up the sigma bond between sb and f in antimony(iii) fluoride, sbf3?

Afina-wow [57]

Answer: The four $sp^3$ hybridized orbitals on Sb makes up the sigma bonds between Sb and F in antimony(iii) fluoride , $SbF_3$

Explanation:

According to VESPR theory:

Number of electrons around the central atom : $\frac{1}{2}[V+N-C+A]$

V = number of valence electrons

N = number of neighboring atoms

C = charge on cation

A = charge on an anion

In antimony(III) fluoride ,

Antimony being central atom: V= 5,N =3,C=0,A=0

Number of electrons : $\frac{1}{2}[V+N-C+A]=4$

Number of electrons around the central atom are 4 which means that $SbF_3$ molecule has four $sp^3$ hybridized orbitals.

6 0

3 years ago

Can anyone help me out please?

Wewaii [24]

Answer:

Qustion 7: 

Answer is → "c. Calcium ions and sulfate ions join together to form an insoluble compound"

Question 8: 

Answer is → b. ²³⁹Pa

Question 9: 

Answer is → "c. 1/3 the pressure in container B"

Explanation:

=================================================================

Qustion 7: 

from the graph:

a. precipitation reactions only occur when solution temperature are decreased.

this option is wrong due to precipitation reaction can occur at room temperature or even at elevated temperature.

b. Sodium ions and chloride ions repel each other when they are in a solution.

Also, this is a wrong statement because sodium ions is positively charged and attract the negatively charged chloride ions.

d. the creation of precipitate adds mass that was previously non-existent.

From graph it is clear that the mass doesn't changed before and after the reaction remain constant (=300.23 g).

So, the right answer is

→ "c. Calcium ions and sulfate ions join together to form an insoluble compound"

=================================================================

Question 8: 

According to the equation :

²³⁹U → ________+ ₋₁⁰e
uranium(239) loses a beta particle and so the mass numer will not change ( remains 239).

So, the right answer is

→ b. ²³⁹Pa

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Question 9: 

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and T are constant, and have two different values of P and V:

Pₐ * Vₐ= P(b) *V(b)

from the graph:

Vₐ= 9.0 L.

V(b)= 3.0 L.

∴ Pₐ = [P(b) * V(b)] / Vₐ = [P(b) * 3.0 L] / 9.0 L = 1/3 P(b).

So, the right choice is:

Answer is → "c. 1/3 the pressure in container B"

=================================================================

3 0

3 years ago

HELP please quick HELP

stiks02 [169]

What do you need help with

4 0

3 years ago

Please- why does my lesson have no information on what's in my assignment. anyways, please help :')

taurus [48]

Answer:

the answer is 55 cg/L

Explanation:

Knowing the conversions, there is 100 cg in one g. if there is .55 g/mL then you multiply by a 100 to get 55 cg/L.

4 0

3 years ago

If 18.1 g of ammonia is added to 27.2 g of oxygen gas, how many grams of excess reactant is remaining once the reaction has gone

GREYUIT [131]

Answer:

m of NH3 = 6.46 g

Explanation:

First, in order to know the limiting and excess reactant, we need to write and balance the equation that is taking place:

NH₃ + O₂ ---------> NO + H₂O

Now, let's balance the equation:

4NH₃ + 5O₂ ---------> 4NO + 6H₂O

Now that we have the balanced equation, let's see which reactant is in excess. To know that, let's calculate the moles of each reactant using the molar mass:

MM NH3 = 17 g/mol

MM O2 = 32 g/mol

moles NH3 = 18.1 / 17 = 1.06 moles

moles O2 = 27.2 / 32 = 0.85 moles

Now, let's compare these moles with the theorical moles that the balanced equation gave:

4 moles NH3 --------> 5 moles O2

1.06 moles ----------> X

X = 1.06 * 5 / 4 = 1.325 moles of O2

These means in order to NH3 completely reacts with O2, it needs 1.325 moles of O2, which we don't have it. We only have 0.85 moles of O2, therefore, the limiting reactant is the O2 and the excess is NH3.

Now, let's see how many grams in excess we have left after the reaction is complete.

4 moles NH3 --------> 5 moles O2

X moles NH3 ----------> 0.85 moles

X = 0.85 * 4 / 5 = 0.68 moles of NH3

This means that 0.85 moles of O2 will react with only 0.68 moles of NH3, and we have 1.06 so, the remaining moles are:

moles remaining of NH3 = 1.06 - 0.68 = 0.38 moles

Finally the mass:

m = 0.38 * 17

m = 6.46 g of NH3

8 0

3 years ago