The reaction of bromine gas with chlorine gas, shown here, has a Keq value of 7.20 at 200°C. If a closed vessel was charged with the two reactants, each at an initial concentration of 0.200 M, but with no initial concentration of BrCl, what would be the equilibrium concentration of Br2, Cl2 and BrCl(g)?
Br2(g) + Cl2(g) ↔ 2BrCl(g) K = 7.20

Answer:

1) B

2) D

3) A

4) Ga

5) K

6)Po

7) Atomic size increases down the group

8) B<Al<Ga<In<Tl

9)Se<C<Ga

10) ionization energy is the energy required to remove electrons from the outermost shell of an atom.

Explanation:

In the periodic table, the properties of elements reoccur ''periodically'' throughout the table, hence the name 'periodic table'.

Ionization energy increases across the period hence the noble gas He has the highest ionization energy.

Since ionization energy increases across the period, group 1 elements possess the lowest ionization energy.

Since atomic size increases down the group and decreases across the period, gallium is smaller than indium, potassium is smaller than caesium, polonium is smaller than titanium and iodine is larger than bromine.

This explanation above justifies the order of increasing atomic radius of group 13 elements shown in answer number 8 above.

Since atomic size decreases across the period, the order of increasing atomic size shown in answer number 9 above is correct.

Ionization energy is the energy required to remove electrons from the outermost shell of an atom.

Answer:

See explanation

Explanation:

The electrophilic substitution of aromatic compounds occurs faster in substituted aromatic compounds due to the fact that the ring becomes more or less susceptible to electrophilic attack depending on the nature of the substituent in the ring.

Electron pushing substituents such as alkyl groups stabilize the positive charge developed during electrophilic substitution hence they activate the ring towards electrophilic substitution.

The methyl group is an ortho - para directing substituent hence the product obtained by nitration of toluene is o-nitrotoluene and p-nitrotoluene.

The stepwise mechanism for obtaining these products is shown in the image attached to this answer.

Answer:

So, look up for electronegativity table

There's a continuum for deciding whether the bond is non-polar, polar, or ionic

If electronegativity difference is less than 0.4 it's non-polar

Explanation:

Pls vote my answer as brainliest

Answer:

compound light microscope

Answer:

C₁₅N₃H₅

Explanation:

Let's assume we have 240 g of the dye (1 mol), in that case we'd have:

Now we convert the masses of each element into moles, using their respective molar masses:

Thus the molecular formula is C₁₅N₃H₅.

Answer:

A) Electrons orbit the nucleus in shells at fixed distances

Answer:

the answer is A

Explanation:

Answer:

0.01684 s⁻¹

Explanation:

In a first-order reaction, the rate is proportional to the concentration of only one reactant (raised by 1). In this case, the reactant is H₂O₂. Thus, the rate law is the following:

rate = k [H₂O₂]

We have the following data for the initial rate:

rate = 0.00842 mol/(L·s)

[H₂O₂] = 0.500 mol/L

So, we introduce the data in the expression for the rate law to calculate k:

k = rate/[H₂O₂] = (0.00842 mol/L·s)/0.500 mol/L = 0.01684 s⁻¹

Answer:

B. There is a very large percentage of C-12.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to realize that, since the average atomic mass is 12.01 amu, then the C-12, with an atomic mass of 12.000 am prevails over C-13 with an atomic mass of 13.003 amu as long as the average is nearer to the former.

In such a way, the answer will be B. There is a very large percentage of C-12.

Regards!

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of [tex]-6.4^oC[/tex] and a molal freezing point depression constant [tex]K_f= 3.96^oC.kg/mol[/tex]. A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at [tex]-13.6^oC[/tex]. Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

Answer: 129.66 g of glycine will be dissolved.

Explanation:

Depression in the freezing point is the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

[tex]\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m[/tex]

                                 OR

[tex]\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times \frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent (g)}}[/tex]        ....(1)

where,

i = Van't Hoff factor = 1 (for non-electrolytes)

Freezing point of pure solvent = [tex]-6.4^oC[/tex]

Freezing point of solution = [tex]-13.6^oC[/tex]

[tex]K_f[/tex] = freezing point depression constant  = [tex]3.96^oC/m[/tex]

[tex]M_{solute}[/tex] = Molar mass of solute (glycine) = 75.07 g/mol

[tex]w_{solvent}[/tex] = Mass of solvent = 950 g

Plugging values in equation 1:

[tex]-6.4-(-13.6)=1\times 3.96\times \frac{\text{Given mass of glycine}\times 1000}{75.07\times 950}\\\\\text{Given mass of glycine}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\\text{Given mass of glycine}=129.66g[/tex]

Hence, 129.66 g of glycine will be dissolved.

Answer:

37.85 L

Explanation:

3.785 x 10.00 = 37.85 L

it would take 37.85 L  to fill a 10.00 tank

(sorry if im wrong pls dont report)

(hope this helps can i plz have brainlist :D hehe)

Answer:

See explanation and image attached

Explanation:

The reaction between benzophenone and phenylmagnesium bromide is a Grignard reaction.

A Grignard reagent is any alkyl magnesium halide compound. In this case, the Grignard reagent is phenylmagnesium bromide.

Reaction of Grignard reagent with a ketone yields all alcohol. Thus, the reaction of benzophenone with phenylmagnesium bromide yields triphenyl methanol.

The mechanism of the reaction and all electron movements are shown in the image attached to this answer.

Answer:

When the negatively-charged rod is brought close to the electroscope, positive charges are attracted to it and negative charges are repelled away from it. The electroscope has a net neutral charge and the rubber rod has a net negative charge. If they are brought into contact, they will both take a net negative charge.

Explanation:

I looked it up

Answer:

A 1.0 g sample of propane, C3H8, was burned in the calorimeter.

The temperature rose from 28.5 0C to 32.0 0C and the heat of combustion 10.5 kJ/g.

Calculate the heat capacity of the calorimeter apparatus in kJ/0C

Explanation:

[tex]Heat of combustion = heat capacity of calorimeter * deltaT\\[/tex]

Given,

The heat of combustion = 10.5kJ/g.

[tex]deltaT = (32.0-28.5)^oC\\deltaT = 3.5^oC[/tex]

Substitute these values in the above formula to get the value of heat capacity of the calorimeter.

[tex]deltaT =heat capacity of calorimeter * (change in temperature)\\10.5kJ/g = heat capacity of calorimeter * (3.5^oC)\\\\=>heat capacity of calorimeter = \frac{10.5kJ/g}{3.5^oC} \\=>heat capacity of calorimeter = 3.0 kJ/g.^oC[/tex]

Answer:

The heat capacity of the calorimeter is [tex]3.0kJ/g.^oC.[/tex]

Answer:

a) 320: two significant figures.

b) 2,366: four significant figures.

c) 73.0: three significant figures.

d. 532.5: four significant figures.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to write each number by knowing we move the decimal places to the right as much as the exponent is, and also, we count every figure, even zeros, because they are to the right of the first nonzero digit:

a) 320: two significant figures because the rightmost zero is not preceded o followed by a decimal place.

b) 2,366: four significant figures.

c) 73.0: three significant figures, because the zero is followed by the decimal place.

d. 532.5: four significant figures.

Regards!

Answer:

Substance 1 will diffuse at a faster rate.

Explanation:

We can solve this problem by keeping in mind Fick's law, which states:

Where:

As (dc/dx) is equal for both substances, as stated by the problem, the substance with the higher diffusion constant will diffuse at a faster rate.

Thus the answer is substance 1.

Answer:

The correct answer is - hard water reacts to form the calcium or magnesium salt of the organic acid of the soap.

Explanation:

Soaps are made up of fatty acids or oils by treating with strong alkali and are salts of sodium and potassium. Hard water, on the other hand, has a high concentration of minerals in comparison to soft water. When hard water and soap are mixed together the salts of the minerals like Ca2+ and Mg2+ ions present in hard water react with fatty acids of the soap.

The sodium salts are changed to the salts of calcium and magnesium which are precipitated to an insoluble substance. The insoluble salts of the calcium or magnesium dirt stick on the clothes thus cleaning ability of soap is reduced.

The correct answer is mixture

Answer:

2.5 × 10² ppm

Explanation:

Step 1: Given data

Step 2: Convert 0.050 g to μg

We will use the conversion factor 1 g = 10⁶ μg.

0.050 g × 10⁶ μg/1 g = 5.0 × 10⁴ μg

Step 3: Calculate the concentration of NaCl in ppm

The concentration of NaCl in ppm is equal to the micrograms of NaCl per gram of the sample.

5.0 × 10⁴ μg NaCl/200. g = 2.5 × 10² ppm

Answer:250 ppm

Explanation:

Answer:

Methanol and butanol are alcohols. Alcohols have the same_______________ as alkanes and the __________ identifies the compound as an alcohol.

Explanation:

Alcohols belong to a group of organic compounds which contain -OH group as the functional group.

So alcohols have the same carbon -hydrogen bonds as alkanes and the        -OH functional group identifies the compound as an alcohol.

Answer:

57.2 g

Explanation:

First we convert 66.4 grams of Ba(ClO₄)₂·3H₂O into moles, using its molar mass:

0.170 moles of Ba(ClO₄)₂·3H₂O would produce 0.170 moles of 0.170 moles of Ba(ClO₄)₂. Meaning we now convert 0.170 moles of Ba(ClO₄)₂ into grams, using the molar mass of Ba(ClO₄)₂:

Answer:

CH3CHO+H2O → CH3OCH3 - addition

CH,CICH CI + Zn → C2H4 + ZnCl2 - elimination

CH3CH3Br + OH – CH3CH3OH + Br - substitution

2CH2COOH >>(CH3CO)20 + H20 - condensation

Explanation:

An addition reaction is a reaction in which a specie is added across the double bond as we can see in CH3CHO+H2O → CH3OCH3.

In an elimination reaction, a small molecule is lost from a saturated compound to form the corresponding unsaturated compound as in CH,CICH CI + Zn → C2H4 + ZnCl2

In a substitution reaction, a chemical moiety replaces another in a molecule as in; CH3CH3Br + OH – CH3CH3OH + Br .

A condensation reaction is in which two molecules are joined together to form a bigger molecule as in; 2CH2COOH >>(CH3CO)20 + H20.

Answer:

addition

elimination

substitution

condensation

Explanation:

[tex]q = 350 \times0.897 \times (70 - 25) \\ q = 14127.75[/tex]

Answer:

wavy, a feel of water, smooth and costy

Explanation:

Answer:

Density = 19.3 g/cm³

Explanation:

In order to answer this question we need to keep in mind the following definition of density:

As both the mass and the volume are given by the problem, we can proceed to calculate the density of gold:

Answer:

Explanation:

As the measurements are given to us in centimeters, let's start by calculating the surface area in square centimeters:

Now we convert 100.75 cm² to m², as follows:

Finally we convert 0.010075 m² to nm², as follows:

Answer:

pH = 3.68

Explanation:

We can solve this problem by using Henderson-Hasselbach's equation:

Assuming we have 1 L of the buffer solution then the molar concentrations of formate and formic acid would be:

We now have all required data to calculate the pH:

Explanation:

Aromatic nitrogen bonded to hydrogen, as found in pyrrole must be represented as [nH] ; thus imidazole is written in SMILES notation as n1c[nH]cc1 . When aromatic atoms are singly bonded to each other, such as in biphenyl, a single bond must be shown explicitly: c1ccccc1-c2ccccc2 .

Answer:

[Ar] 4s²3d³

Explanation:

Vanadium has atomic number of 23. The electronic configuration of vanadium can be written as:

V (23) => 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d³

NOTE: After the 18th electron, 4s will be filled before 3d.

We can also write the electronic configuration of an element in its condensed form by writing the symbol of the noble before the desired element in a squared bracket followed by the remaining electrons to complete the electronic configuration of the element.

The electronic configuration of vanadium in its condensed form is given below:

The noble gas before vanadium is Argon (Ar) with atomic number of 18. Thus, the electronic configuration of vanadium becomes:

V (23) => [Ar] 4s²3d³