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3 years ago

Identify the type of the bond

Chemistry

1 answer:

julsineya [31]3 years ago

8 0

That is phosphorus trichloride and a COVALENT BOND

Cl needs one electron and P has metallic character, it donates electron easy

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How many moles of na are needed to produce 4 moles of nacl in this reaction? 2na + cl2 → 2nacl?

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14...........................

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3 years ago

consider this reaction at equilibrium at a total pressure: 2h2o(g) o2(g) 2h2o2(g) suppose the volume of this system is twice its

Daniel [21]

The total pressure when the new equilibrium is stabilized is half of the initial pressure of the system.

The given chemical reaction at a stable equilibrium is,

2H₂O(g)+O₂(g) = 2H₂O₂(g)

According to the ideal gas equation,

PV = nRT

P is pressure,

V is volume,

n is moles

R is gas constant,

T is temperature.

Assuming the temperature is constant.

If the volume of the system is twice the initial volume then the total pressure at the new equilibrium can be found out as,

P₁V₁ = P₂V₂

Where, P₁ and V₁ are initial volume and pressure while P₂ and V₂ are final pressure and volume.

If V₂ = 2V₁,

P₂ = P₁/2

So, the final total pressure will be half of the initial pressure.

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11 months ago

Which of the following gases would be most likely to experience ideal behavior at high pressures?

Delvig [45]

D, Monotonic gases, which have no inter molecular attractions are most suited as ideal gases

4 0

3 years ago

Read 2 more answers

The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t

slava [35]

Answer: The theoretical yield of iron(III) sulfate is 26.6 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol$

The given chemical equation follows:

$2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4$

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = $\frac{1}{2}\times 0.133=0.0665moles$ of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

$0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g$

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

8 0

3 years ago

A sample gas initially occupies 3.35

Strike441 [17]

After i don’t get it

7 0

3 years ago

Identify the type of the bond​

Identify the type of the bond