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stepladder [879]
2 years ago
5

The reaction between alcohols and carboxylic acids is called

Chemistry
1 answer:
aleksandrvk [35]2 years ago
4 0

Explanation:

When a carboxylic acid is treated with an alcohol and an acid catalyst, an ester is formed (along with water). This reaction is called the Fischer esterification.

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Help meee, if u help me i give u a pizza :)
kompoz [17]
Give me 25 minutes and I’ll have the answers :)
6 0
2 years ago
A sample of a compound made entirely from copper and oxygen was found to have a total mass of 0.143 g the mass of copper in the
tensa zangetsu [6.8K]

Mass of Oxygen: 0.0159 grams

Moles of Oxygen: 9.94x10^-4

To find the mass of oxygen, subtract the mass of copper from the total mass.

0.143-0.1271=0.0159

There are 0.0159 grams of Oxygen.

To find how many moles there are, divide the given amount of oxygen by the molar mass (atomic mass) of oxygen because that mass is the same as one mole of oxygen.

Molar mass of Oxygen: 16.00

0.0159/16.00=9.94*10^{-4}

There are 9.94*10^-4 moles of Oxygen.

6 0
1 year ago
How many moles of steam are produced when 5.74 moles of octane (C3H18) react?
jonny [76]

Answer:

51.66 mol H2O (steam)

Explanation:

5.74 mol C3H18 x 18 mol H2O/ 2 mol C3H18 = 51.66 mol H2O (steam)

6 0
2 years ago
Do all colors of light have the same energy?
icang [17]
No, darker colors like purple and blue have higher wave frequency which transmit more energy.
6 0
3 years ago
If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W
olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3

Therefore, the leftover of sodium nitrate is:

m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3

Finally, the percent yield is computed via:

Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%

Best regards!

6 0
2 years ago
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