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Alik [6]
3 years ago
13

AlF3 is almost insoluble in anhydrous HF but dissolves if KF is present.passage of BF3 through the resulting solution causes AlF

3 to precipitate... can anyone explain the idea behind this?
Chemistry
1 answer:
miskamm [114]3 years ago
4 0
<span>It's because the product formed with BF3 is more complex which able to decompose AlF3.

</span>AlF3 doesn't dissolve in HF because of the fluorine. It's doesn't allow for coordination due to the hydrogen. However, it will dissolve in KF. If you look at the chemical reaction, it's able to form a salt.

<span>3KF+Al<span>F3</span>−>3KF.Al<span>F3</span></span>
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The spontaneous reaction that occurs when the cell in the picture operates is as follows: 2Ag+ + Cd(s) ???? 2 Ag(s) + Cd2+ (A) V
Snowcat [4.5K]

Answer:

14. B    15. D    16. C     17. B

Explanation:

The spontaneous reaction that occurs when the cell operates is shown below:

2Ag^{+} + Cd_{(s)} ⇒2Ag_{(s)} + Cd^{2+}

We need to select the correct option from the list below for the following questions.

(A) Voltage increases. (B) Voltage decreases but remains > zero. (C) Voltage becomes zero and remains at zero. (D) No change in voltage occurs. (E) Direction of voltage change cannot be predicted without additional information.

14. A 50-milliliter sample of a 2-molar Cd(NO_{3})_{2} solution is added to the left beaker.

If a 50-milliliter sample of a 2-molar Cd(NO_{3})_{2}  solution is added to the left beaker, the voltage decreases but its value remains greater than zero. The correct option is B

15. The silver electrode is made larger.

If the silver electrode is made larger, no change in the value of the voltage since we don't have the idea of the initial value. The correct option is D.

16. The salt bridge is replaced by a platinum wire.

If the salt bridge is replaced by a platinum wire, there will be no passage of electrons because electrons can't pass through a platinum wire. Therefore, the voltage will be zero and remains at zero. The correct option is C.

17. Current is allowed to flow for 5 minutes.

If current is allowed to flow for 5 minutes, the voltage decreases but its value remains greater than zero. The correct option is B.

7 0
3 years ago
The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of B
loris [4]

I believe here is the right question, so will just ignore the rest of the junk information from the previous message

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.944 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of BHT

Answer:

C_{15}H_{24}0

Explanation:

A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.994 g of H2O(g).

If all the carbon in BHT is present in CO_2 and also, all the hydrogen in BHT is  present in H_2O, Then we can determine for the corresponding numbers of moles of Carbon(C) and Hydrogen (H) respectively as:

moles of  CO_2 = 8.990 g*(\frac{1mole}{44.01g})

                       =  0.2043 moles

∴ moles of C =  0.2043 moles

moles of H_2O = 2.944 g *(\frac{1mole}{18.01g} )

                       = 0.1635 moles

∴ moles of H = 2 × 0.1635 moles

                      = 0.327 moles

Since number of moles= \frac{mass}{molarmass}

number of moles of H =  0.327 moles

molar mass of H = 1.008 g/mol

∴  mass of H in the sample = 0.327 moles × 1.008 g/mol

                                             = 0.329616g

                                             

mass of C in the sample can be calculated as = 0.2043 moles × (\frac{12.01g}{1 mole} )

= 2.453643 g

mass of C+H in the sample = 2.453643g + 0.329616g

= 2.783259 g

mass of O can be calculated as = 3.001 g - 2.783259 g

= 0.217741 g

∴ moles of O = 0.217741g × (\frac{1mole}{16.0g})

= 0.0136 moles

Now, since; we've gotten our data, we can now proceed to calculate for the empirical formula.

C                                          H                                    O

0.2043                                0.327                             0.0136

Dividing by the least number (0.0136) , we have :

\frac{0.2043}{0.0136}                                     \frac{0.327}{0.0136}                               \frac{0.0136}{0.0136}

15.02                                      24.04                             1

≅

15                                             24                                  1

Therefore, the empirical formula would be : C_{15}H_{24}0

7 0
2 years ago
The effectiveness of nitrogen fertilizers depends on both their ability to deliver nitrogen to plants and the amount of nitrogen
densk [106]

Answer:

Ammonia > Urea > Ammonium nitrate > Ammonium sulphate

Explanation:

Percentage by mass of nitrogen in NH3:

Molar mass of NH3= 17 g/mol

Hence % by mass = 14/17 × 100 = 82.35%

% by mass of NH4NO3

Molar mass of NH4NO3 = 80.043 g/mol

Hence; 28/80.043 × 100 = 34.98%

% by mass of (NH4)2SO4;

Molar mass of (NH4)2SO4= 132.14 g/mol

Hence; 28/132.14 × 100 = 21.19%

% by mass of CH4N2O

Molar mass of urea = 60.0553 g/mol

Hence 28/60.0553 × 100 = 46.62%

8 0
3 years ago
How do you Calculate the molarity of 250 ml of solution in which 2.7 g of MgCl2 are dissolved
IRINA_888 [86]

Answer:

Molar mass of MgCl2 is 95 g/mol

Mg = 24 g/mol and Cl = 35.5 ×2 = 71 g/mol

moles = mass given/ molar mass

= 2.7/95 = 0.028 mol

volume = 250/1000 = 0.25 dm3 (ml is the same as dm3)

molarity of MgCl2 = moles/volume

= 0.028/0.25

= 0.112 mol/dm3

3 0
2 years ago
How to do part (b)??
laila [671]

it has less tightly bound electrons, is able to lose electron easily as compare to metal B at it has 4 unpaired electron in 3d sub-shell.

3 0
3 years ago
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