All categories

3 years ago

Whats the molar volume (of a solution)?

Chemistry

1 answer:

il63 [147K]3 years ago

7 0

<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3

Answer:

Molarity can be used to calculate the volume of solvent or the amount of solute. The relationship between two solutions with the same amount of moles of solute can be represented by the formula c1V1 = c2V2, where c is concentration and V is volume.

<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3

You might be interested in

Sabendo que A Densidade do Metal De Chumbo e 11,3/cm3

a_sh-v [17]

Sabendo que A Densidade do Metal De Chumbo e 11,3/cm3 means Knowing that the density of lead Metal and 11,3/cm3.

3 0

4 years ago

How does things go sublime

Lemur [1.5K]

Things go sublime when changing automatically into vapor when it is heated, usually forming a solid figure.

5 0

3 years ago

Group I element are very active.give reason

RUDIKE [14]

Answer:

wat

Explanation:

7 0

3 years ago

Un medicamento corresponde a una disolución de "nafazolina" (sustancia activa) en agua purificada, en una concentración de 12% V

ratelena [41]

Answer:

12 mL

Explanation:

Según la pregunta;

V / V% concentración del fármaco = 12%

volumen de la solución = 100 ml

De;

V / V% = volumen de fármaco / volumen de solución

volumen de fármaco = V / V% × volumen de solución

volumen de fármaco = 12/100 × 100

= 12 mL

5 0

3 years ago

An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i

lutik1710 [3]

Answer:

$C_{metal}=126.6\frac{J}{g\°C}$

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

$Q_{metal}=-Q_{water}$

Which can be also written as:

$m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})$

Thus, since we need the specific heat of the metal, we solve for it as shown below:

$C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}$

Best regards.

7 0

3 years ago