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yawa3891 [41]
3 years ago
15

The strongest forces of attraction occur between molecules of(1) HCl (3) HBr(2) HF (4) HI

Chemistry
1 answer:
Alexxandr [17]3 years ago
4 0
It would likely be HF. It displays hydrogen bonding. 
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The following data were measured for the reaction BF3(g)+NH3(g)→F3BNH3(g): Experiment [BF3](M) [NH3](M) Initial Rate (M/s) 1 0.2
Rama09 [41]

Answer:

-r_{A}=k\times[BF_3]^{1}\times[NH_3]^{1}

Explanation:

The rate law of a chemical reaction is given by

-r_{A}=k\times[BF_3]^{\alpha}\times[NH_3]^{\beta}

This law can be written for any experiment, and making the quotient between those expressions the reaction orders can be found

Between experiments 1 and 2  

\frac{-r_{A1}}{{-r}_{A2}}=\left(\frac{\left[NH_3\right]_1}{\left[NH_3\right]_2}\right)^\beta

Then the expression for the calculation of \beta

\beta=\frac{ln\frac{-r_{A1}}{-r_{A2}}}{ln\left(\frac{\left[NH_3\right]_1}{\left[NH_3\right]_2}\right)}=\frac{ln\frac{0.2130}{0.1065}}{ln\left(\frac{0.250}{0.125}\right)}

Resolving  

\beta=1

Doing the same between experiments 3 and 4 the expression for \alpha is

\alpha=\frac{ln\frac{-r_{A3}}{-r_{A4}}}{ln\left(\frac{\left[BF_3\right]_3}{\left[BF_3\right]_4}\right)}=\frac{ln\frac{0.0682}{0.1193}}{ln\left(\frac{0.200}{0.350}\right)}

Resolving  

\alpha=1

This means that the rate law for this reaction is  

-r_{A}=k\times[BF_3]^{1}\times[NH_3]^{1}

5 0
3 years ago
What type of energy is carried by electrical charges as they move in a circuit?
Usimov [2.4K]
B. Electrical energy
3 0
3 years ago
What is Avagrados Number?
dsp73
6.02x10^23; This represents the number of molecules in 1 mole of a substance.
8 0
3 years ago
Read 2 more answers
What is the value for AG at 100 Kif AH = 27 kJ/mol and AS = 0.09 kJ/(mol-K)?
Elina [12.6K]

Answer:

ΔG =   18KJ/mol

Explanation:

Given data:

ΔS = 0.09 Kj/mol.K

ΔH = 27 KJ/mol

Temperature  = 100 K

ΔG = ?

Solution:

Formula:

ΔG = ΔH - TΔS

ΔH = enthalpy

ΔS = entropy

by putting values,

ΔG =  27 KJ/mol - 100K(0.09 Kj/mol.K)

ΔG =  27 KJ/mol - 9 KJ/mol

ΔG =   18KJ/mol

7 0
2 years ago
Assume you find four bottles in an empty laboratory, each containing a liquid. The labels that were on these bottles have fallen
fenix001 [56]

Answer:

No, I can not identify the contents of each bottle using solubility and polarity (with H2O) information

Explanation:

While it is true that polar substances dissolve in water and nonpolar substances do not dissolve in water, the task here is to specifically identify the contents of each of the bottles.

Solubility in water can not tell us exactly what liquid is which substance. For instance, trans-1,2-dichloroethylene, cis-1,2-dichloroethylene and cyclooctane are all insoluble in water. The fact that they do not dissolve in water does not tell us which liquid is which compound.

Even though acetic acid is miscible with water, it is not a conclusive prove that the liquid is acetic acid since other polar organic compounds are also miscible in water.

It is only by determining the boiling point of each substance that I can conclusively identify the contents of each bottle since boiling point is an intrinsic property of substances.

6 0
2 years ago
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