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EastWind [94]
3 years ago
8

What is the element for Trifluoroborane

Chemistry
1 answer:
Harlamova29_29 [7]3 years ago
7 0

<span>Ethoxyethane; trifluoroborane; BF3.Et2O; Boron trifluoride ethyl ether; Boron trifluoride diethyl ether; Boron trifluoride-diethyl ether; Boron 

</span>
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How many moles of NH3 can be produced from 30.0 mol of H2 and excess N2?
VMariaS [17]
20 mol of NH, can be produce from 30 mol o H2
6 0
4 years ago
The reaction 2HgO (s)→2Hg (I)+O2 (g) has a percent yield of 50%. You want to produce 100 g of Hg.
nevsk [136]

The mass of HgO needed for the reaction is 216 g

The correct answer to the question is Option C. 216 g

We'll begin by calculating the theoretical yield of Hg.

  • Actual yield of Hg = 100 g
  • Percentage yield = 50%
  • Theoretical yield of Hg =?

Theoretical yield = Actual yield / percentage yield

Theoretical yield = 100 / 50%

Theoretical yield of Hg = 200 g

Finally, we shall determine the mass of HgO needed for the reaction.

2HgO → 2Hg + O₂

Molar mass of HgO = 201 + 16 = 217 g/mol

Mass of HgO from the balanced equation = 2 × 217 = 434 g

Molar mass of Hg = 201 g/mol

Mass of Hg from the balanced equation = 2 × 201 = 402 g

From the balanced equation above,

402 g of Hg were produced from 434 g of HgO.

Therefore

200 g of Hg will be produce by = (200 × 434) / 402 = 216 g of HgO.

Thus, 216 g of HgO is needed for the reaction.

Learn more about stoichiometry:

brainly.com/question/24426334

4 0
2 years ago
Read 2 more answers
What is the oxidation number of Pt in K₂PtCl₆?
Agata [3.3K]

Answer:

the oxidation number of Pt in K₂PtCl₆ <u>is 4.</u>

K₂PtCl₆=0

2+x-6=0

x=4

8 0
3 years ago
Help me answer this plz.
boyakko [2]

Answer:

87.58 L of C₂H₂

Explanation:

We'll begin by calculating the number of mole in 250 g of CaC₂.

This can be obtained as follow:

Mass of CaC₂ = 250 g

Molar mass of CaC₂ = 40 + (12×2)

= 40 + 24

= 64 g/mol

Mole of CaC₂ =?

Mole = mass /Molar mass

Mole of CaC₂ = 250 / 64

Mole of CaC₂ = 3.91 moles

Next, the balanced equation for the reaction. This is given below:

CaC₂ + 2H₂O —> C₂H₂ + Ca(OH)₂

From the balanced equation above,

1 mole of CaC₂ reacted to produce 1 mole of C₂H₂.

Next, we shall determine the number of mole of C₂H₂ produced by the reaction of 250 g (i.e 3.91 moles) of CaC₂. This can be obtained as follow:

From the balanced equation above,

1 mole of CaC₂ reacted to produce 1 mole of C₂H₂.

Therefore, 3.91 moles of CaC₂ will also react to produce 3.91 moles of C₂H₂.

Finally, we shall determine the volume of C₂H₂ produced from the reaction. This can be obtained as follow:

Recall:

1 mole of any gas occupy 22.4 L at STP.

1 mole of C₂H₂ occupied 22.4 L at STP.

Therefore, 3.91 moles of C₂H₂ will occupy = 3.91 × 22.4 = 87.58 L

Thus, 87.58 L of C₂H₂ is produced from the reaction.

5 0
3 years ago
A polymer P is made up of two monodisperse fractions; fraction A with molecular weight of 1000 g/mole and fraction B with a mole
vesna_86 [32]

Answer:

a)Number average molecular weight is 50, 500 g/mol

b) Weight average molecular weight is 99, 019.8 g/mol

Explanation:

We have a polymer P made up of two monodisperse fractions.

A with molecular weight of MA = 1000 g/mol and B with MB =100000 g/mol.  

The batch contains an equal mole fraction of each component A and B.  

Let's suppose a total number (Nt) of mols 2 moles. Equal fraction means XA = 0.5 and XB =0.5

Nt = 2 mol

Na = 2*0.5 = 1 mol

Nb = 2*0.5 = 1 mol.

So, we have 1 mol of A, 1 mol of B and 2 moles in total.  

a) The number average molecular weight (NAM) is calculate using the mole numbers of each component. In this case, we will multiple each component molecular weight by the number of moles of each one. After that we will sum them and finally to divide by the total number of moles.

NAM = (Na*MA + Nb*MB)/(Nt)

NAM = (1 mol *1000 g/mol + 1*100000 g/mol ) /(2 mol)

NAM = 50500 g/mol

The number average molecular weight for the polymer P is 50,500 g/mol

b) Weight average molecular weight (WAM) is calculated using the mass quantities of each component. Weight mass of A (WA), weight mass of B (WB) are calculate using the moles of A, B and their molecular weights respectively. Total Weight (WT)

WA = Na*MA = 1 mol *1000 g/mol = 1000 g A

WB = Nb*MB = 1mol * 100000 g/mol = 100 000 gB

WT = WA + WB = 101 000 g

Now we will calculate average molecular using weights, we will multiple each component molecular weight by the mass of each one. After that we will sum them and finally to divide by the total mass.

WAM = (WA*MA + WB*MB)/(WT)

WAM = (1000 g *1000 g/mol + 100000 g*100000 g/mol )/(101 000 g)

WAM = 99 019.8 g/mol

The weight average molecular weight for polymer P is 99, 019.8 g/mol

6 0
4 years ago
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