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2 years ago

8

The class of organic compound below contains a carbonyl group as a part of its structure?

1 answer:

kolbaska11 [484]2 years ago

5 0

Aldehydes and Ketones. Aldehydes and ketones are classes of organic compounds that contain a carbonyl (C=O) group.

that's the correct answer

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Which is the largest?<br> O 43.32 mL<br> 0 0,0056 L<br> O 0.000076 KL

mel-nik [20]

Answer:

43.32 ML is bigger

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3 years ago

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Find the final equilibrium temperature when 15.0 g of milk at 13.0 degrees c is added to 148 g of coffee with a temperature of 8

user100 [1]

According to zeroth law of thermodynamics, when two objects are kept in contact, heat (energy) is transferred from one to the other until they reach the same temperature (are in thermal equilibrium). When the objects are at the same temperature there is no heat transfer.

So, at equilibrium, $q_{lost}$ = $q_{gain}$ , $q_{lost}= q_{milk}$ + $q_{coffee}$

q=m×c×T, where q = heat energy, m = mass of a substance, c = specific heat (units J/kg∙K), T is temperature

$q_{lost}= (15X13X4.19)+(148X88.3X4.19), (15+148)X4.19XT_{final}$ =(15X13X4.19)+(148X88.3X4.19)

$T_{final}$ = 81.37 ° C

6 0

3 years ago

Help meeeeeeeeeeeeeee

lesya [120]

AMSWER :

CHEMICAL FORMULA :

• WATER - H2O

• CARBONDIOXIDE - CO2

• SODIUM HYDROXIDE - NaOH

• SULPHURIC ACID - H2SO4

• SODIUM CHLORIDE - NaCI

• SODIUM BICARBONATE - NaCHO3

8 0

2 years ago

What is the mass or 1.53 x 10^23 molecules of H2SO4?

Kaylis [27]

Answer:

24.9 g H2SO4

Explanation:

1.53 x 10^23 / 6.02x10^23 * 98.079g = 24.9 g

3 0

3 years ago

If 5.12 g of oxygen O2 gas occupies a volume of 6.21L at a certain temperature and pressure, how many grams of oxygen gas will o

ddd [48]

Answer : The mass of $O_2$ occupy 30.3 L under the same conditions will be, 24.9 grams.

Explanation :

First we have to calculate the moles of $O_2$

$\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}=\frac{5.12g}{32g/mol}=0.16mol$

Now we have to calculate the moles of $O_2$ in 30.3 L by using Avogadro's law.

Avogadro's law : It is defined as the volume of gas is directly proportional to the number of moles of gas at constant pressure and temperature.

$V\propto n$

or,

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

where,

$V_1$ = initial volume of gas = 6.21 L

$V_2$ = final volume of gas = 30.3 L

$n_1$ = initial moles of gas = 0.16 mol

$n_2$ = final temperature of gas = ?

Now put all the given values in the above equation, we get:

$\frac{6.21L}{0.16mol}=\frac{30.3L}{n_2}$

$n_2=0.781mol$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

Molar mass of $O_2$ = 32 g/mol

$\text{ Mass of }O_2=(0.781moles)\times (32g/mole)=24.9g$

Therefore, the mass of $O_2$ occupy 30.3 L under the same conditions will be, 24.9 grams.

6 0

3 years ago