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pochemuha
3 years ago
11

The compound, CH3COOH, is an example of which type of organic compound?

Chemistry
1 answer:
sattari [20]3 years ago
3 0
Hi,
CH3COOH is a carboxylic acid since it contains a carboxyl group (-COOH). The general formula of a carboxylic acid would be R-COOH. CH3COOH is know as acetic acid or ethanoic acid.
Hope this helps! If my answer was not clear enough or you’d like further explanation please let me know.
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A sample of an unknown gas at STP has a density of 0.630 gram per liter. What is the gram molecular mass of this gas?
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<span>STP means standard temperature and pressure at 0°C (273K) and 1 atm (atmosphere). The density of the unknown gas is 0.63 gram per liter. The deal gas equation is PV = nRT. The n is the numer of moles and can be represented as mass of the gas, m, divided by the molar mass, c.  so we have,</span>  

PV = nRT
PV = (m/c)RT
Since the density is d = m/V
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Pc = dRT
c = drT/P  

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3 years ago
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3 years ago
Consider the reaction 3X + 2Y 5C + 4D
xeze [42]

Answer:

This problem is providing a chemical equation between two hypothetical elements, X and Y and asks for the molesof X that are needed to 

produce 21.00 moles of D in excess Y. After the following work, the answer turns out to be 15.75 mol X:Mole ratios:In chemistry, one the most crucial branches is stoichiometry, which allows us to perform calculations with grams, moles and particles (atoms, molecules and ions). It is based on the concept of mole ratios, whereby the moles of a specific substance can be converted to moles of another one, say product to reactant, reactant to reactant, reactant to product and product to product.

Calculations:In such a way, since 21.00 moles of D are given, we need the mole ratio of D to X in order to get the answer, which according to the reaction is 3:4 based on their coefficients in the reaction. Hence, we calculate the required as follows:

Explanation:

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5 0
2 years ago
A sample of gallium Bromide GaBr2,weighing 0.165 g was dissolved in water and treated with silver nitrate AgNO3, and resulting t
tresset_1 [31]

<u>Answer:</u> The percent gallium in gallium bromide is 30.30 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of gallium bromide = 0.165 g

Molar mass of titanium gallium bromide = 229.53 g/mol

Putting values in equation 1, we get:

\text{Moles of gallium bromide}=\frac{0.165g}{229.53g/mol}=0.00072mol

  • The chemical equation for the reaction of gallium bromide and silver nitrate follows:

GaBr_2+2AgNO_3\rightarrow 2AgBr(s)+Ga(NO_3)_2

By Stoichiometry of the reaction:

1 moles of gallium bromide produces 1 mole of gallium nitrate

So, 0.00072 moles of gallium bromide will produce = \frac{1}{1}\times 0.00072=0.00072moles of gallium nitrate

  • Now, calculating the mass of gallium nitrate from equation 1, we get:

Molar mass of gallium nitrate = 193.73 g/mol

Moles of gallium nitrate = 0.00072 moles

Putting values in equation 1, we get:

0.00072mol=\frac{\text{Mass of gallium nitrate}}{193.73g/mol}\\\\\text{Mass of gallium nitrate}=0.139g

Calculating the mass of gallium in the reaction, we use unitary method:

In 1 mole of gallium nitrate, 1 mole of gallium atom is present.

In 193.73 grams of gallium nitrate, 69.72 g of gallium atom is present.

So, in 0.139 grams of gallium nitrate, the mass of gallium present will be = \frac{69.72}{193.73}\times 0.139=g

  • To calculate the percentage composition of gallium in gallium bromide, we use the equation:

\%\text{ composition of gallium}=\frac{\text{Mass of gallium}}{\text{Mass of gallium bromide}}\times 100

Mass of gallium bromide = 0.165 g

Mass of gallium = 0.050 g

Putting values in above equation, we get:

\%\text{ composition of gallium}=\frac{0.050g}{0.165g}\times 100=30.30\%

Hence, the percent gallium in gallium bromide is 30.30 %.

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The answer is:

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