All categories

3 years ago

Pro∆G0=-RT in k Known as

Chemistry

1 answer:

stepan [7]3 years ago

6 0

Answer:

Gibbs free energy equation

Explanation:

You might be interested in

All but one statement below tell why scientists work in groups

USPshnik [31]

I think it’s D hope that helped

6 0

3 years ago

If 5.0 grams of sucrose, C12H22O11, are dissolved in 10.0 grams of water, what will be the boiling point of the resulting soluti

GaryK [48]

Answer : The boiling point of the resulting solution is, $100.6^oC$

Explanation :

Formula used for Elevation in boiling point :

$\Delta T_b=i\times k_b\times m$

or,

$T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^oC$

$k_b$ = boiling point constant = $0.52^oC/m$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute (sucrose) = 5.0 g

$w_1$ = mass of solvent (water) = 10.0 g

$M_2$ = molar mass of solute (sucrose) = 342.3 g/mol

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

$(T_b-100)^oC=1\times (0.52^oC/m)\times \frac{(5.0g)\times 1000}{342.3\times (10.0g)}$

$T_b=100.6^oC$

Therefore, the boiling point of the resulting solution is, $100.6^oC$

5 0

3 years ago

A red sports drink contains Red 40 dye. A 5.4 mL aliquot of this sports drink was diluted to 25.0 mL with deionized water in a v

miskamm [114]

Answer:

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

Explanation:

Concentration of red dye in sport drink before dilution $C_1=?$

Volume of the sport drink before dilution $V_1=5.4 mL$

Concentration of red dye in sport drink after dilution $C_2=18.1 ppm$

Volume of the sport drink after dilution $V_2=25.0 mL$

$C_1V_1=C_2V_2$ ( dilution )

$C_1=\frac{C_2V_2}{V_1}=\frac{18.1 ppm\times 25.0 mL}{5.4 mL}$

$C_1=83.80 ppm\approx 84.0ppm$

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

6 0

3 years ago

4. Plants use energy from sunlight, water, and carbon dioxide to produce sugar. Which structure is found only in plant cells and

professor190 [17]

The answer is C chloroplast

7 0

3 years ago

If 155 J of heat energy is applied to a block of silver weighing 27.9 g, by how many degrees will the temperature of the silver

Brilliant_brown [7]

Answer:

The temperature of silver increase by 23.15 °C.

Explanation:

Given data:

Heat added = 155 j

Mass of silver = 27.9 g

How many degree temperature increase = ?

Solution:

Specific heat capacity:

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

Specific heat capacity of silver is = 0.240 j/g.°C

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

Now we will put the values in formula.

155 j = 27.9 g × 0.240 j/g.°C × ΔT

155 j = 6.696 j/°C × ΔT

ΔT = 155 j /6.696 j/°C

ΔT = 23.15 °C

3 0

3 years ago

Pro∆G0=-RT in k Known as​

Pro∆G0=-RT in k Known as