if a gas sample has a pressure of 30.7 kPa at 273 K, what does the temperature have to decrease to lower the pressure to 28.4 kPa?​
253K
268K
262K

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:

Explanation:

There are 7 protons in nitrogen

7 TYPE OF NITRONGEN

. . .

Answer:

im not sure

Explanation:

i went on here looking nobody answered yet

Answer:

3.7 x 10²¹ O₂ molecules

Explanation:

2KClO₃ => 2KCl + 3O₂

given 0.500g KClO₃ => 0.500g/122.55g/mol = 0.0041 mole KClO₃

0.0041 mole KClO₃ => 3/2(0.0041) mole O₂ = 0.0061 mole O₂

0.0061 mole O₂ = 0.0061 mole O₂ x 6.02 x 10²³ molecules O₂/mole O₂

= 3.7 x 10²¹ molecules O₂

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:

C

Explanation:

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

Answer:

B the weight of the passengers

Answer:

Dominant

Explanation:

hope this help

Answer:

They refer to energy that moves. Kinetic energy is energy that is currently moving. Potential refers to energy that has yet to move, or simply energy in wait

Explanation:

Answer:

I'm not 100% but a possible answer could be: No energy could either be created or destroyed

LMK if not but good luck!

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:

A)Renewable Energy Sources

Explanation:

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

Explanation:

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.

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:

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 !

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:

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.

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:

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

Answer: [tex]2NH_3(g)+2O_2(g)\rightarrow N_2O(g)+3H_2O(l)+684kJ[/tex]

Explanation:

The skeletal thermochemical equation for the reaction is:

[tex]NH_3(g)+O_2(g)\rightarrow N_2O(g)+H_2O(l)+342kJ[/tex]

The balanced  thermochemical equation for the reaction is:

[tex]2NH_3(g)+2O_2(g)\rightarrow N_2O(g)+3H_2O(l)+684kJ[/tex]

When 1 mole of [tex]NH_3[/tex] reacts with oxygen , heat released = 342 kJ

Thus when 2 moles of [tex]NH_3[/tex] reacts with oxygen , heat released = [tex]\frac{2}{1}\times 342 kJ=684kJ[/tex]

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:

B-100

Explanation:

ppm is an unit of concentration that could be defined as the mass in mg of solute (In this case, NaCl) per kg of solution.

Now, a solution of NaCl that is 0.01% by mass, contains 0.01g of NaCl in 100g of solution.

To solve this question, we must convert the mass of NaCl to mg and the mass of solution to kg:

Mass NaCl:

0.01g * (1000mg / 1g) = 10mg

Mass Solution:

100g * (1kg / 1000g) = 0.10kg

The ppm are:

10mg / 0.10kg =

100ppm

Right answer is:

Answer:

newtons 3rd law of motion

Explanation:

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!

Explanation:

pdrias darme la traduccion no te entiendo

Answer:

The answer is "Option b and Option c".

Explanation:

This buffer is a buffer of ammonia and ammonium ion. Thus it requires the solution [tex]NH_3 \ \ and \ \ NH_4^{+}[/tex].

In point 1:

The solution containing [tex]NH_3 \ \ and \ \ NH_4^{+}[/tex] at 1M concentration would be given by mixing the two solutions. Thus, this buffer is a legitimate route.

In point 2:

It gives the ions you want but they are not the same.

In point 3:

[tex]1 M[/tex] [tex]NH_3[/tex] and [tex]1 M[/tex][tex]HCl[/tex] volume would not produce the same [tex]NH_3 \ \ and \ \ NH_4^{+}[/tex] concentrations. Therefore, this buffer isn't a valid route.

In point 4:

Some [tex]1 M[/tex] [tex]NH_3[/tex] volume and half [tex]HCl[/tex] . This offers the same rate as half.

Answer:

here you are

Explanation:

atomic energy