What word is used synonymously with the word "base"

How many grams of baf2 are in 3.7 moles of baf2

Mice21 [21]

To convert moles to gram multiply 3.7 moles of BaF2 to the molar mass of BaF2

BaF2 molar mass = 175.326

3.7 moles x 175.326g = 648.70g BaF2

8 0

4 years ago

Predict the carboxylic acid product of the following reaction:CH3COOCH2CH3ethyl ethanoate+H2OH+⟶carboxylic acid+CH3CH2OHethanolE

GaryK [48]

Answer: The IUPAC name of the carboxylic acid formed is ethanoic acid.

Explanation:

The basic rules for naming of organic compounds are :

First select the longest possible carbon chain.
The longest possible carbon chain should include the carbons of double or triple bonds.
The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne by adding the suffix -yne and carboxylic acid by adding the suffix -oic acid.
The numbering is done in such a way that first carbon of double or triple bond gets the lowest number.
The carbon atoms of the double or triple bond get the preference over the other substituents present in the parent chain.

For the given chemical reaction, the equation follows:

$CH_3COOCH_2CH_3+H_2O\rightarrow CH_3COOH+CH_3CH_2OH$

When hydrolysis of ester takes place, it leads to the formation of an alcohol and a carboxylic acid.

So, when hydrolysis of ethyl ethanoate occurs, it produces ethanol and a 2- Carbon carboxylic acid named as ethanoic acid.

Hence, the IUPAC name of the carboxylic acid formed is ethanoic acid.

8 0

4 years ago

The system co2(g) + h2(g) ⇀↽ h2o(g) + co(g) is at equilibrium at some temperature. at equilibrium a 4.00 l vessel contains 1.00

Marina CMI [18]

Answer: The moles of $CO_2$ added to the system is 7.13 moles

Explanation:

We are given:

Moles of $CO_2$ at equilibrium = 1.00 moles

Moles of $H_2$ at equilibrium = 1.00 moles

Moles of $H_2O$ at equilibrium = 2.40 moles

Moles of $CO$ at equilibrium = 2.40 moles

Volume of the container = 4.00 L

Concentration is written as:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

The given chemical equation follows:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}$

Putting values in above equation, we get:

$K_c=\frac{(\frac{2.40}{4.00})\times (\frac{2.40}{4.00})}{(\frac{1.00}{4.00})\times (\frac{1.00}{4.00})}\\\\K_c=5.76$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of CO = 0.791 mol/L

Volume of solution = 4.00 L

Putting values in above equation, we get:

$0.791M=\frac{\text{Moles of CO}}{4.00L}\\\\\text{Moles of CO}=(0.791mol/\times 4.00L)=3.164mol$

Extra moles of CO = (3.164 - 2.40) = 0.764 moles

Let the moles of $CO_2$ needed be 'x' moles.

Now, equilibrium gets re-established:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

Initial: 1.00 1.00 2.40 2.40

At eqllm: (0.236+x) 0.236 3.164 3.164

Again, putting the values in the expression of $K_c$ , we get:

$5.76=\frac{(\frac{3.164}{4.00})\times (\frac{3.164}{4.00})}{(\frac{0.236+x}{4.00})\times (\frac{0.236}{4.00})}\\\\5.76=\frac{10.011}{0.056+0.236x}\\\\x=7.13$

Hence, the moles of $CO_2$ added to the system is 7.13 moles

4 0

3 years ago

Which of these following terms are made of matter and which or not made of matter: air,light,radio,magnetic field,car,radio wave

viva [34]

Matter: radio, car, flashlight, textbook, human

Not matter: everything besides the matter items

3 0

3 years ago

potassium chlorate (kclo3) decomposes into potassium chloride (kcl) and oxygen gas (o2) how many grams of oxygen can be produced

sergeinik [125]

Answer:

$m_{O_2}=354.24gO_2$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$

In such a way, as 7.38 moles of potassium chloride are decomposed, the resulting grams of oxygen are computed considering a 2 to 3 molar relationship in the chemical reaction:

$m_{O_2}=7.38molKClO_3*\frac{3molO_2}{2molKClO_3}*\frac{32gO_2}{1molO_2} \\\\m_{O_2}=354.24gO_2$

Best regards.

7 0

4 years ago

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