A student dissolves 10.7 g of lithium chloride (LiCl) in 300. g of water in a well-insulated open cup. He then observes the temperature of the water rise from 22.0 °C to 28.6 °C over the course of 3.8 minutes. Use this data, and any information you need from the ALEKS Data resource:
LiCl(s) rightarrow Li+(aq) + Cl-(aq)
You can make any reasonable assumptions about the physical properties of the solution. Note for advanced students: it's possible the student did not do the experiment carefully, and the values you calculate may not be the same as the known and published values for this reaction.
1) Is this reaction exothermic, endothermic, or neither?
2) If you said the reaction was exothermic or endothermic, calculate the amount of heat that was released or absorbed by the reaction in this case.
3) Calculate the reaction enthalpy delta Hrxn per mole of LiCl.

Answer:

[tex]\boxed {\boxed {\sf 2.76 \ mol}}[/tex]

Explanation:

Molarity is found by dividing the moles of solute by liters of solution.

[tex]molarity = \frac {moles}{liters}[/tex]

We know the molarity is 1.2 M (mol\liter) and there are 2.3 liters of solution. Substitute the known values into the formula.

[tex]1.2 \ mol/liter= \frac {x}{2.3 \ liters}[/tex]

Since we are solving for x, we must isolate the variable. It is being divided by 2.3 and the inverse of division is multiplication. Multiply both sides by 2.3 liters.

[tex]2.3 \ liters *1.2 \ mol/liter= \frac {x}{2.3 \ liters}* 2.3 \ liters\\2.3 *1.2 \ mol= x\\2.76 \ mol =x[/tex]

In a solution with a molarity of 1.2 and 2.3 liters of solution, there are 2.76 moles.

Answer: The answer is a. limiting, excess. I took the test.

Explanation:

Answer:

newtons 3rd law of motion

Explanation:

The data set is missing in the question. The data set is given in the attachment.

Solution :

a). In the table, there are four positive examples and give number of negative examples.

Therefore,

[tex]$P(+) = \frac{4}{9}$[/tex]   and

[tex]$P(-) = \frac{5}{9}$[/tex]

The entropy of the training examples is given by :

[tex]$ -\frac{4}{9}\log_2\left(\frac{4}{9}\right)-\frac{5}{9}\log_2\left(\frac{5}{9}\right)$[/tex]

= 0.9911

b). For the attribute all the associating increments and the probability are :

  [tex]$a_1$[/tex]   +   -

  T   3    1

  F    1    4

Th entropy for   [tex]$a_1$[/tex]  is given by :

[tex]$\frac{4}{9}[ -\frac{3}{4}\log\left(\frac{3}{4}\right)-\frac{1}{4}\log\left(\frac{1}{4}\right)]+\frac{5}{9}[ -\frac{1}{5}\log\left(\frac{1}{5}\right)-\frac{4}{5}\log\left(\frac{4}{5}\right)]$[/tex]

= 0.7616

Therefore, the information gain for [tex]$a_1$[/tex]  is

  0.9911 - 0.7616 = 0.2294

Similarly for the attribute [tex]$a_2$[/tex]  the associating counts and the probabilities are :

  [tex]$a_2$[/tex]  +   -

  T   2    3

  F   2    2

Th entropy for   [tex]$a_2$[/tex] is given by :

[tex]$\frac{5}{9}[ -\frac{2}{5}\log\left(\frac{2}{5}\right)-\frac{3}{5}\log\left(\frac{3}{5}\right)]+\frac{4}{9}[ -\frac{2}{4}\log\left(\frac{2}{4}\right)-\frac{2}{4}\log\left(\frac{2}{4}\right)]$[/tex]

= 0.9839

Therefore, the information gain for [tex]$a_2$[/tex] is

  0.9911 - 0.9839 = 0.0072

[tex]$a_3$[/tex]     Class label      split point       entropy        Info gain

1.0         +                        2.0            0.8484        0.1427

3.0        -                         3.5            0.9885        0.0026

4.0        +                        4.5            0.9183         0.0728

5.0        -

5.0        -                        5.5            0.9839        0.0072

6.0        +                       6.5             0.9728       0.0183

7.0        +

7.0        -                        7.5             0.8889       0.1022

The best split for [tex]$a_3$[/tex]  observed at split point which is equal to 2.

c). From the table mention in part (b) of the information gain, we can say that [tex]$a_1$[/tex] produces the best split.

Answer:

B the weight of the passengers

Answer:

here you are

Explanation:

atomic energy

Answer:

1. 1.25 mol ants x 6.02*10^23 ants/1 mol ants = 7.53*10^23 ants

2. 4.92*10^26 pencils x 1 mol pencils/6.02*10^23 pencils = 817 mol pencils

3. 0.26 mol molecules x 6.02*10^23 molecules/1 mol molecules = 1.6*10^23 molecules

4. 3.46*10^19 molecules x 1 mol molecules/6.02*10^23 molecules = 5.75*10^-5 mol molecules

5. 5.3*10^20 atoms x 1 mol atoms/6.02*10^23 atoms = 8.8 mol atoms

6. 0.11 mol atoms x 6.02*10^23 atoms/1 mol atoms = 6.6*10^22 atoms

I would suggest looking into "dimensional analysis" for help with this type of material. Dimensional analysis will stick with you all throughout chemistry, so picking it up will be extremely beneficial.

Explanation:

pdrias darme la traduccion no te entiendo

Answer:

C. gravity causing a curved path as an object tries to go straight.

Answer:

A)Renewable Energy Sources

Explanation:

Answer:

65.1 grams

Explanation:

The percentage of impurities in the contaminated CaCl₂ sample is:

Meaning that the purity of the sample is (100 - 7) 93%.

Now we calculate the mass of pure CaCl₂ in the contaminated sample:

Thus the answer is 65.1 grams, as that amount is how much CaCl₂ is in the sample and it would be impossible to obtain more than that.

Answer:

b

Explanation:

Answer:

C

Explanation:

It is run by Elon Musk, who is determined to get to Mars.

Answer: The percentage of potassium iodide in the sample is 2.63 %.

Explanation:

The chemical equation for the reaction of iodide ions with bromine gas follows:

[tex]I^-+3Br_2+3H_2O\rightarrow 6Br^-+IO_3^-+6H^+[/tex]                (i)

The chemical equation for the reaction of iodate ions with barium ions follows:

[tex]Ba^{2+}+2IO_3^-\rightarrow Ba(IO_3)_2[/tex]                 ......(ii)

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given mass of barium iodate = 0.0596 g

Molar mass of barium iodate = 487.13 g/mol

Using equation 1:

[tex]\text{Moles of barium iodate}=\frac{0.0596 g}{487.13 g/mol}\\\\\text{Moles of barium iodate}=1.22\times 10^{-4} moles[/tex]

By Stoichiometry of the reaction (ii):

1 mole of barium iodate is produced by 2 moles of iodate ions

So, [tex]1.22\times 10^{-4} moles[/tex] of barium iodate will be produced by [tex]\frac{2}{1}\times 1.22\times 10^{-4} =2.44\times 10^{-4}moles[/tex] of iodate ions

By the stoichiometry of the reaction (i):

1 mole of iodate ions are produced by 1 moles of iodine ions

So, [tex]2.44\times 10^{-4} moles[/tex] of iodate ions will be produced by [tex]\frac{1}{1}\times 2.44\times 10^{-4} =2.44\times 10^{-4}moles[/tex] of iodine ions

Moles of potassium iodide = Moles of iodide ions = [tex]2.44\times 10^{-4}[/tex]

Since, the molar mass of potassium iodide = 166 g/mol

Using equation 1:

[tex]\text{Mass of potassium iodide}=2.44\times 10^{-4}mol\times 166 g/mol\\\\\text{Mass of potassium iodide}=0.0405 g[/tex]

To calculate the percentage by mass of a substance, the equation used is:

[tex]\text{Percent by mass}=\frac{\text{Mass of a substance}}{\text{Mass of solution}}\times 100[/tex]

Mass of a solution = 1.54 g

Mass of potassium iodide = 0.0405 g

Using above equation:

[tex]\text{Percent potassium iodide}=\frac{0.0405 g}{1.54g}\times 100\\\\\text{Percent potassium iodide}=2.63\%[/tex]

Hence, the percentage of potassium iodide in the sample is 2.63 %.

Expressing the results of potassium iodide in percentage = 2.63%

The chemical reaction of iodine ions with Bromine gas can be expressed as :

I⁻  + 3Br₂  + 3H₂O  -- > 6Br⁻ + IO₃ + 6H⁺ ----- ( 1 )

Chemical reaction between iodate ions with barium ions  can be expressed as :  Ba²⁺  + 2IO⁻₃ ------> Ba ( IO₃ )₂   --------- ( 2 )

Step 1 : Calculate the number of Barium iodate moles

mass of Barium iodate = 0.0596 g

molar mass of Barium iodate = 487.13 g/mol

from equation ( 1 )

moles of Barium iodate = ( 0.0596 ) / ( 487.13 ) = 1.22 * 10⁻⁴ moles

also from equation ( 1 ) the moles of potassium iodide = moles of iodide ions

= 2.44 * 10⁻⁴

molar mass of potassium iodide = 166 g/mol

Next step : Determine the mass of potassium iodide

moles of potassium * molar mass

= 2.44 * 10⁻⁴  * 166 g/mol  = 0.0405 g

Final step : Determine the percentage of potassium iodide in the solution

Percentage = ( mass of potassium iodide / mass of solution ) * 100

                   = ( 0.0405 / 1.54 ) * 100

                   = 2.63%

Hence we can conclude that potassium iodide in percentage = 2.63%

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Answer:

The value of n is 6

Explanation:

The half-reactions of the problem are:

Cr(s) → Cr³⁺ + 3e⁻

Pb²⁺ + 2e⁻→ Pb(s)

To balance the electrons we must multiply the half-reactions as follows:

2 * (Cr(s) → Cr³⁺ + 3e⁻)

3 * (Pb²⁺ + 2e⁻→ Pb(s))

2Cr(s) → 2Cr³⁺ + 6e⁻

3Pb²⁺ + 6e⁻→ 3Pb(s)

In Nernst equation, the value of n are the electrons used to balance the reaction, as in this problem, the electrons are 6:

Answer:

im not sure

Explanation:

i went on here looking nobody answered yet

Answer:

A saturated solution is a mixture in which the maximum amount of a given solute has been dissolved into the solvent. ... At this point adding more solute will not change the concentration of the solution; adding more solute will simply result in more solid at the bottom of the solution.

Answer:

will not change

Explanation:

A saturated solution is a mixture in which the maximum amount of a given solute has been dissolved into the solvent. ... At this point adding more solute will not change the concentration of the solution; adding more solute will simply result in more solid at the bottom of the solution.

Answer:

due to the pollution the light months tended to stand out against the the wood that was dark due to the smoke the darker months population grew due to the fact that they were more harder to see for the predators

Answer:

The dark colored months became more common

Explanation:

Thats the answer !

Answer:

4 Al + 3 O2 → 2 Al2O3

(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al

(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2

0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.

((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =

10.1 g O2 left over

Explanation:

i really hope this is what you were looking for <3 :)

Answer:

Since the equation for amount remaining is (1/2)^n, where n stand for the number of half lives, after 2.3 half lives the equation is (1/2)^2.3. This makes the answer about   1/4.92, which converts to about 20% of the original amount.

Explanation:

Since the equation for amount remaining is (1/2)^n, where n stand for the number of half lives, after 2.3 half lives the equation is (1/2)^2.3. This makes the answer about   1/4.92, which converts to about 20% of the original amount.

Answer:

The correct answer is

- after deprotonation carboxylic acid stabilizes by binding -OH group with hydrogen bonding in salicylic acid and while it's not possible with its constitutional isomer (para-hydroxy benzoic acid).

The more electronegative atoms are replaced by the H atom of the carboxylic group and form more stabilize carboxylic acid which not takes place in para-hydroxy benzoic acid.

Molar mass of C3H8 = C 3 (12.01 g/mol) = 36.03 (g/mol)

H 8 (1.008 g/mol) = 8.064 (g/mol)

44.09 (g/mol)

74.6 g propane x 1 mole propane x 6.022 x 10

23

molecules

44.09 g propane 1 mole propane

= 1.02 x 10

24 molecules propane

Answer:

reactivity

Explanation:

for example atoms in group 7 react by gaining 1 electron to become stable but they do not have the same number of energy levels

Answer:

Bond order is a counting method that gives an idea about numbers of electrons shared between atoms. A species with a higher bond order is more stable. A bond order equal to 2 is a double bond, and a bond order of 3 is a triple bond.

Explanation:

Answer:

0.150M

Explanation:

hope it helps!!!!!!!!!!!!

The molarity of the resulting solution when 300. mL of a 0.400 M solution is diluted to 800 mL is 0.15M.

Molarity is defined as the ratio of the number of moles of solute present in a solution, to the volume of that particular solution in litres.

It is calculated by using the formula

M₁V₁ = M₂V₂

M₁= initial molarity of the solution

V₁ = initial volume of the solution

M₂ = final molarity of the solution

V₂ = final volume of the solution

Given data,

M₁ = 0.400 M (initial molarity)

V₁ = 300 mL (initial volume)

V₂ = 800 mL (final volume)

Substituting the values we get

M₁V₁ = M₂V₂

.4× 300/800 = M₂

Therefore, molarity of the resulting solution = .15

To know more about molarity here

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Answer:

[tex]74.5gCO_2[/tex]

Explanation:

Hello there!

In this case, for the described reaction, it is possible to realize there is a 13:8 mole ratio between oxygen (O2) and carbon dioxide (CO2); moreover, since the molar mass of the former is 32.00 g/mol and that of latter is 44.01 g/mol, the produced mass of the required product turns out to be:

[tex]88gO_2*\frac{1molO_2}{32.00gO_2}*\frac{8molCO_2}{13molO_2}*\frac{44.01gCO_2}{1molCO_2} \\\\=74.5gCO_2[/tex]

Best regards!

Answer:

Trees

Explanation:

I think sorry if I get it wrong