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djverab [1.8K]
2 years ago
15

Identify the type of reaction: 2AgCl + BaBr2 -> 2AgBr + BaCl2

Chemistry
1 answer:
Marta_Voda [28]2 years ago
6 0

Answer:

C

Explanation:

Just look at the reactants.

2AgCl + BaBr2

The first reactant is made of 2 elements.

The second reactant is made of 2 elements.

It can't be a decomposition. At this level there is only one reactant made of 2 elements. Something like

2MgO ===> 2Mg + O2

is a decomposition.  One compound breaking down into 2 elements Mg and O.

It can't be a combustion. One of the reactants in a combustion is oxygen. Those equations look like

C3H8 + 5O2 ==> 3CO2 + 4H2O

That would be what a combustion looks like

It can't be a single replacement. They look like

Mg + CuO ===> Cu + MgO

There are elements on both sides of the reaction.

that leaves a double replacement which I wrote about how you distinguish it.

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P-fluoroanisole reacts with sulfur trioxide and sulfuric acid. Draw the major product of this substitution reaction; if applicab
Ipatiy [6.2K]

Answer: (Structure attached).

Explanation:

This type of reaction is an aromatic electrophilic substitution. The overall reaction is the replacement of a proton (H +) with an electrophile (E +) in the aromatic ring.

The aromatic ring in p-fluoroanisole has two sustituents, an <u>halogen</u> and a <u>methoxy group</u>, which are <em>ortho-para</em> directing substituents.

Aryl sulfonic acids are easily synthesized by an electrophilic substitution reaction aromatic using <u>sulfur trioxide as an electrophile</u> (very reactive).

The reaction occurs in three steps:

  1. The attack on the electrophile forms the sigma complex.
  2. The loss of a proton regenerates an aromatic ring.
  3. The sulfonate group can be protonated in the presence of a strong acid (H₂SO₄).

Normally, a mixture of <em>ortho-para</em> substituted products would be obtained. However, since both <em>para</em> positions are occupied, only the <em>ortho </em>substituted product is obtained here.

8 0
3 years ago
What is the theoretical yield of aspirin ( C 9 H 8 O 4 ), which has a molar mass of 180.15 g/mol, possible when reacting 3.03 g
pantera1 [17]

Answer:

The theoretical yield of aspirin is 3.95 grams

Explanation:

Step 1: Data given

Mass of salicylic acid = 3.03 grams

Volume of acetic anhydride = 3.61 mL

Density of acetic anhydride = 1.08 g/cm³

Step 2: The balanced equation

C4H6O3+C7H6O3→C9H8O4+C2H4O2

Step 3: Calculate moles salicylic acid

Moles salicylic acid = mass salicylic acid / molar mass salicylic acid

Moles salicylic acid = 3.03 grams /138.121 g/mol

Moles salicylic acid = 0.0219 moles

Step 4: Calculate mass acetic anhydride

Mass acetic anhydride = volume * density

Mass acetic anhydride = 3.61 mL * 1.08 g/mL

Mass acetic anhydride = 3.90 grams

Step 5: Calculate moles acetic anhydride

Moles acetic anhydride = 3.90 grams / 102.09 g/mol

Moles acetic anhydride = 0.0382 moles

Step 6: Calculate limiting reactant

For 1 mol salicylic acid we need 1 mol acetic anhydride to produce 1 mol aspirin

Salicylic acid is the limiting reactant. It will completely be consumed. (0.0219 moles). Acetic anhydride is in excess. There will react 0.0219 moles. There remain 0.0382 - 0.0219 =0.0163 moles

Step 7: Calculate moles aspirin

For 1 mol salicylic acid we need 1 mol acetic anhydride to produce 1 mol aspirin

For 0.0219 moles salicylic acid we'll have 0.0219 moles aspirin

Step 8: Calculate theoretical yield of aspirin

Mass of aspirin = moles aspirin *molar mass aspirin

Mass of aspirin = 0.0219 moles *180.15 g/mol

Mass of aspirin = 3.95 grams

The theoretical yield of aspirin is 3.95 grams

7 0
3 years ago
An unknown amount of acid can often be determined by adding an excess of base and then back-titrating the excess. A 0.3471−g sam
MAXImum [283]

Explanation:

The given data is as follows.

       Mass of mixture = 0.3471 g

As the mixture contains oxalic acid and benzoic acid. So, oxalic acid will have two protons and benzoic acid has one proton.

This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.

Hence,   moles of NaOH in 97 ml = \frac{0.1090 \times 97}{1000}

                                                       = 10.573 \times 10^{-3} mol

Moles of HCl in 21.00 ml = \frac{0.2060 \times 21}{1000}

                                         = 4.326 \times 10^{-3} mol

Therefore, total moles of NaOH that reacted are as follows.

           10.573 \times 10^{-3} mol - 4.326 \times 10^{-3} mol      

                = 6.247 \times 10^{-3} mol

So, total 3 mole of NaOH will react with 1 mole of mixture. Therefore, number of moles of NaOH reacted with benzoic acid is as follows.

                  \frac{6.247 \times 10^{-3}}{3}

                    = 2.082 \times 10^{-3} mol

Since, molar mass of NaOH is 40 g/mol. Therefore, calculate the mass of NaOH as follows.

                    2.082 \times 10^{-3} mol \times 40 g/mol

                         = 83.293 \times 10^{-3} g

                         = 0.0832 g

Whereas molar mass of benzoic acid is 122 g/mol.

Therefore,       40 g NaOH = 122 g benzoic acid

So,            0.0832 g NaOH = \frac{122 g}{40 g} \times 0.0832 g

                                             = 0.253 g

Hence, calculate the % mass of benzoic acid as follows.

                      \frac{0.253 g}{0.3471 g} \times 100

                           = 73.10%

Thus, we can conclude that mass % of benzoic acid is 73.10%.

4 0
3 years ago
I WILL GIVE BRAINLIEST PLEASE HELP ME I NEED HELP!
AleksAgata [21]
I think the answer would be trenches but I’m sorry if I’m wrong
4 0
3 years ago
If you have a solution that contains 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH, what is the mole fraction of
irakobra [83]

Answer:

0.22

Explanation:

Given, Mass of C_{18}H_{21}NO_3 = 46.85 g

Molar mass of C_{18}H_{21}NO_3 = 299.4 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{46.85\ g}{299.4\ g/mol}

Moles\ of\ C_{18}H_{21}NO_3= 0.1565\ mol

Given, Mass of C_{2}H_{5}OH = 125.5 g

Molar mass of C_{2}H_{5}OH = 46.07 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{125.5\ g}{46.07\ g/mol}

Moles\ of\ C_{2}H_{5}OH= 0.5535\ mol

So, according to definition of mole fraction:

Mole\ fraction\ of\ codeine=\frac {n_{codeine}}{n_{codeine}+n_{ethanol}}

Mole\ fraction\ of\ codeine=\frac{0.1565}{0.1565+0.5535}=0.22

4 0
3 years ago
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