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Ray Of Light [21]
2 years ago
9

What is the bond between phosphorus and chlorine?

Chemistry
1 answer:
stepladder [879]2 years ago
5 0

Phosphorus and chlorine form bond is covalent bond.

A covalent bond is likely to form if the difference in electronegativity between the elements is less than 1.7.

For instance, the electronegativity of phosphorus (P) is 2.1 and that of chlorine (Cl) is 3.0.

These two numbers are differentiated by 0.9. Since 0.9 is lesser than 1.7, phosphorus and chlorine form a covalent bond.

Chlorine and phosphorus are both found on the right side of the periodic table, which is where nonmetals are. As a result, a covalent bond will be formed between the two nonmetals.

So, we can say that covalent bond is between phosphorus and chlorine.

To learn more about Phosphorus-chlorine bonds, here :

brainly.com/question/2051818?referrer=searchResults

#SPJ4

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B) Chlorine

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A gas occupies 525 mL at a pressure of 45.0 kPa. What would the volume of the gas be at a pressure of 65.0 kPa
choli [55]

Answer:

The volume of the gas at a pressure of 65.0 kPa would be 363 mL

Explanation:

Boyle's Law is a gas law that relates the pressure and volume of a certain amount of gas, without temperature variation, that is, at constant temperature.

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. In other words, the product P · V remains constant at the same temperature:

P*V=k

Being P1 and V1 the pressure and volume in state 1 and P2 and V2 the pressure and volume in state 2 are fulfilled:

P1*V1=P2*V2

In this case:

  • P1= 45 kPa= 45,000 Pa (being 1 kPa=1,000 Pa)
  • V1= 525 mL= 0.525 L (being 1 L=1,000 mL)
  • P2= 65 kPa= 65,000 Pa
  • V2= ?

Replacing:

45,000 Pa* 0.525 L= 65,000 Pa*V2

Solving:

V2=\frac{45,000 Pa* 0.525 L}{65,000 Pa}

V2=0.363 L=363 mL

<u><em>The volume of the gas at a pressure of 65.0 kPa would be 363 mL</em></u>

6 0
3 years ago
Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)
cupoosta [38]

Answer:

m_{CaCO_3}=0.179gCaCO_3

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3

Regards.

3 0
4 years ago
If kb for nx3 is 1.5×10−6, what is the poh of a 0.175 m aqueous solution of nx3?
Charra [1.4K]
Answer is: pOH = 3,29.
Kb(NH₃) = 1,5·10⁻⁶.
c(NH₃) = 0,175M.
pOH = ?
Chemical reaction: NH₃ + H₂O ⇄ NH₄⁺ + OH⁻.
Kb = c(NH₄⁺) · c(OH⁻) ÷ c(NH₃).
c(NH₄⁺) = c(OH⁻) = x.
x² = Kb · c(NH₃) 
x² = 1,5·10⁻⁶ · 0,175 = 2,625 ·10⁻⁷.
x = c(OH⁻) = √2,625 ·10⁻⁷ = 5,12 · 10⁻⁴.
pOH = -log(c(OH⁻)) = 3,29.

3 0
3 years ago
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