All categories

3 years ago

How many grams are in 23.6 molecules of water?

Chemistry

1 answer:

Akimi4 [234]3 years ago

4 0

Answer:

7.06 g H2O

Explanation:

23.6 molec H2O x $\frac{1 mol H2O}{6.022(10^{23}) }$ x $\frac{18.02 g H2O}{1 mol H2O}$ = 7.06 g H2O

You might be interested in

What describes the behavior of an ideal gas?

valina [46]

Answer:

The answer is D

Explanation:

for a p e x

5 0

3 years ago

Copper has a specific heat of 0.385 J/gºC.

Anna71 [15]

Answer:

The final temperature is 348.024°C.

Explanation:

Given data:

Specific heat of copper = 0.385 j/g.°C

Energy absorbed = 7.67 Kj (7.67×1000 = 7670 j)

Mass of copper = 62.0 g

Initial temperature T1 = 26.7°C

Final temperature T2 = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

Q = m.c. ΔT

7670 J = 62.0 g × 0.385 j/g °C ×( T2- 26.7 °C )

7670 J = 23.87 j.°C ×( T2- 26.7 °C )

7670 J / 23.87 j/°C = T2- 26.7 °C

T2- 26.7 °C = 321.324°C

T2 = 321.324°C + 26.7 °C

T2 = 348.024°C

The final temperature is 348.024°C.

6 0

4 years ago

What are the factors affecting qualitative analysis in chemistry

Kitty [74]

Answer:

Techniques and Tests

Qualitative analysis typically measures changes in color, melting point, odor, reactivity, radioactivity, boiling point, bubble production, and precipitation.

5 0

2 years ago

Match the following with their correct molecular weight. 2-butanone Propyl acetate 4-methyl-2-pentanone Butyl acetate Methanol E

Hatshy [7]

Answer:

2-butanone = 72.11 g/mol (option F)

Propyl acetate = 102.13 g/mol (option C)

4-methyl-2-pentanone = 100.16 g/mol (option D)

Butyl acetate = 116.16 g/mol (option B)

Methanol = 32.04 g/mol (option E)

Ethanol = 46.07 g/mol (option A)

Explanation:

Step 1: Data given

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16.00 g/mol

Step 2:

2-butanone = C4H8O

⇒ 4*12.01 + 8*1.01 + 16.00 = 72.11 g/mol (option F)

Propyl acetate = C5H10O2

⇒ 5*12.01 + 10*1.01 + 2*16.00 = 102.13 g/mol (option C)

4-methyl-2-pentanone = C6H12O

⇒ 6*12.01 + 12*1.01 + 16.00 = 100.16 g/mol (option D)

Butyl acetate = C6H12O2

⇒ 6*12.01 + 12*1.01 + 2*16.00 = 116.16 g/mol (option B)

Methanol = CH3OH = CH4O

⇒ 12.01 + 4*1.01 + 16.00 = 32.04 g/mol (option E)

Ethanol = C2H5OH = C2H6O

⇒ 2*12.01 + 6*1.01 + 16.00 = 46.07 g/mol (option A)

4 0

3 years ago

A una mezcla de 300g, formada con 60% P/P de Hierro y 40% P/P de Arena, se le adicionan 135g de Cobre y 2,77g de Aluminio. ¿Cuál

Svetllana [295]

Answer:

$\%P/P_{hierro}=41.1\%\\\\\%P/P_{arena}=24.4\%\\\\\%P/P_{cobre}=30.8\%\\\\\%P/P_{aluminio}=0.6\%$

Explanation:

¡Hola!

En este caso, dado que estamos tratando con problem sobre porcentaje peso/peso de hierro, arena, cobre y aluminio, primero debemos calcular la masa inicial de estos dos primeros en la mezcla original de acuerdo con:

$m_{hierro}=300g*0.60=180g\\\\m_{arena}=300*0.40=120g$

Ahora si podemos calcular la masa de la mezcla final como la suma de las masas de todos los constituyentes de la mezcla:

$m_T=180g+120g+135g+2.77g=437.77g$

Finalmente, podemos calcular los porcentajes P/P como se muestra a continuación:

$\%P/P_{hierro}=\frac{180g}{437.77g} *100\%=41.1\%\\\\\%P/P_{arena}=\frac{120g}{437.77g} *100\%=24.4\%\\\\\%P/P_{cobre}=\frac{135g}{437.77g} *100\%=30.8\%\\\\\%P/P_{aluminio}=\frac{g}{437.77g} *100\%=0.6\%$

¡Saludos!

5 0

3 years ago