All categories

2 years ago

Is CH3COOCH3 an acid or base?

Chemistry

2 answers:

rodikova [14]2 years ago

7 0

CH3COOCH3 It’s an Acid.

Romashka-Z-Leto [24]2 years ago

4 0

Answer:

acid

Explanation:

Methyl acetate, also known as MeOAc, acetic acid methyl ester or methyl ethanoate, is a carboxylate ester with the formula CH3COOCH3. It is a flammable liquid with a characteristically pleasant smell reminiscent of some glues and nail polish removers.

You might be interested in

A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. what is the partial pressure o

klasskru [66]

Hello!

We have the following statement data:

Data:
$P_{Total} = 800 mmHg$
$P\% N_{2} = 60\%$
$P\% O_{2} = 40\%$
$P_{partial} = ? (mmHg)$

As the percentage is the mole fraction multiplied by 100:

 $P = X_{ O_{2} }*100$

The mole fraction will be the percentage divided by 100, thus:
What is the partial pressure of oxygen in this mixture?

$X_{ O_{2} } = \frac{P}{100}$
$X_{ O_{2}} = \frac{40}{100}$
$\boxed{X_{ O_{2}} = 0.4}$

To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:

In relation to $O_{2}$ :

$\frac{P O_{2} }{P_{total}} = X_O_{2}$
$\frac{P O_{2} }{800} = 0.4$
$P_O_{2} = 0.4*800$
$\boxed{\boxed{P_O_{2} = 320\:mmHg}}\end{array}}\qquad\quad\checkmark$

Answer:
b. 320.0 mm hg

7 0

3 years ago

The number of oxygen atoms in 48.0g of oxygen gas

Doss [256]

Answer:

Now u have 48 g of O2. There. Fore mole=weight/M. W. Of oxygen. Therefor 3mole.

After that if we to multiply the avogadro number with it. So 3 *NA

Now u want only atom calculation then we have 2 molecule of oxygen then multiply it with 2 too.

So final claculation is =3*2*NA.

Explanation:

your welcome

brainlest PLEASEEEEEEE!

3 0

3 years ago

Calculate: (a) the weight (in lbf) of a 30.0 lbm object. (b) the mass in kg of an object that weighs 44N. (c) the weight in dyne

belka [17]

Answer:

a) 965,1 lbf

b) 4,5 kg

c) 1,33 * 10^6 dynes

Explanation:

Mass of an object refers to the amount of mattter it cotains, it can be expressed it gr, kg, lbm, ton, etc.

Weight of an object refers to a force, and is the measurement of the pull of gravitiy on an object. It may be definide as the mass times the acceleration of gravity.

w=mg

In Planet Earth, the nominal "average" value for gravity is 9,8 m/s² (in the International System) or 32,17 ft/s² (in the FPS system).

To solve this problem we'll use the following conversion factors:

1 lbf = 1 lbm*ft/s²

1 N = 1 kg*m/s²

1 dyne = 1 gr*cm/s² and 1 N =10^5 dynes

1 ton = 907,18 kg

1 k = 1000 gr

a) m = 30 lbm

$w = 30 lbm * 32,17 ft/s^{2} = 965, 1 \frac{lmb*ft}{s^{2} } = 965,1 lbf$

b) w = 44 N

First, we clear m of the weight equation and then we replace our data.

$m = \frac{w}{g} = \frac{44 N}{9,8 \frac{m}{{s}^{2}} } = 4,5 kg$

c) m = 15 ton

$m=15 ton * \frac{907,18 kg}{1 ton} = 13607,7 kg \\ w = mg = 13607,7 kg * 9,8 m/s2 = 133355,5 N * \frac{10^{5} dynes }{1 N} = 1,33 * 10^{6}dynes$

4 0

4 years ago

A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. What is the weight/

lara31 [8.8K]

Answer:

A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. The weight/weight % or percent by mass of the solute is :

2.67%

Explanation:

Note : Look at the density of potassium nitrate in water if given in the question.

You are calculating weight /Volume not weight/weight % or percent by mass of the solute

Here the weight/weight % or percent by mass of the solute is asked : So first convert the VOLUME OF SOLUTION into MASS

Density of potassium nitrate in water KNO3 = 2.11 g/mL

$density=\frac{mass}{volume}$

Density = 2.11 g/mL

Volume of solution = 86.4 mL

$2.11=\frac{mass}{86.4}$

$mass = 2.11\times 86.4$

$mass=182.3grams$

Mass of Solute = 4.87 g

Mass of Solution = 183.2 g

w/w% of the solute =

$= \frac{mass\ of\ solute}{mass\ of\ solution}\times 100$

$=\frac{4.87}{183.2}\times 100$

w/w%=2.67%

8 0

3 years ago

How many grams of copper (II) hydroxide can be prepared from 2.4 grams of copper (II) nitrate (Cu(NO3)2 ) and excess sodium hydr

andreyandreev [35.5K]

Answer:

percentage yield = 67%

Explanation:

Mass of Cu(NO₃)₂ = 15.25 g

Mass of NaOH = 12.75 g

Percentage yield = ?

Solution:

Cu(NO₃)₂ + 2NaOH → Cu(OH)₂ + 2NaNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 15.25 g /187.56 g/mol

Number of moles = 0.08 mol

Moles of NaOH :

Number of moles = mass/ molar mass

Number of moles = 12.75 g / 40 g/mol

Number of moles = 0.32 mol

Now we will compare the moles of Cu(OH)₂ with NaOH and Cu(NO₃)₂. NaOH : Cu(OH)₂

2 : 1

0.32 : 1/2×0.32 = 0.16 mol

Cu(NO₃)₂ : Cu(OH)₂

1 : 1

0.08 : 0.08

The number of moles produced by Cu(NO₃)₂ are less so it will limiting reactant.

Mass of Cu(OH)₂:

Mass = number of moles × molar mass

Mass = 0.08 mol × 97.6 g/mol

Mass = 7.808 g

Theoretical yield = 7.808 g

Percent yield:

percentage yield = Actual yield/ theoretical yield × 100

percentage yield = 5.23 g/ 7.808 g × 100

percentage yield = 0.67 × 100

percentage yield = 67%

5 0

3 years ago