How many grams of a solution that is 5.2% sucrose by mass are needed to obtain 18g of sucrose

Answer:

Complete Question:  

Equimolar quantities of CH3OH(l) and C2H5OH(l) are placed in separate 2.0 L containers that have been evacuated beforehand. Pressure gauges are attached to each container, and the temperature is maintained at 300 K. In both containers, liquid is consistently visible at the bottom. The varying pressure within the vessel that contains CH3OH(l) is illustrated below.

In comparison to the equilibrium vapor pressure of CH3OH(l) at 300 K, the equilibrium vapor pressure of C2H5OH(l) at 300 K is

ANSWER : lower, since the London dispersion forces among C2H5OH molecules surpass those among CH3OH molecules.

Explanation:

To clarify the answer provided, let’s begin by defining some concepts.

The London dispersion force is the least strong type of intermolecular force. It is a temporary force that arises when the electron arrangement in two neighboring atoms creates transient dipoles.  

The vapor pressure of a liquid reflects the equilibrium pressure of its vapor above the liquid (or solid); specifically, it represents the pressure associated with the evaporation of a liquid (or solid) in a sealed environment above the substance.

The pressure will be lower due to the stronger London dispersion forces acting between C2H5OH molecules compared to those between CH3OH molecules. This implies that when intermolecular forces are stronger, they intensify the interactions binding the substance together, thereby reducing the liquid's vapor pressure at any given temperature and making it more difficult to vaporize the substance.

Note: The London dispersion force for C2H5OH is more substantial than for CH3OH because C2H5OH has more electrons than CH3OH.

Answer:

Explanation:

Given data:

Initial temperature T₁ = 25.2°C = 298.2K

Initial pressure P₁ = 0.6atm

Final temperature = 72.4°C = 345.4K

What we need to find:

Final pressure = ?

To determine this, we apply a modified version of the combined gas law with constant volume. This simplifies our calculations to:

Here, P and T signify pressure and temperatures, 1 refers to initial and 2 to final temperatures.

Now we can substitute the known variables:

P₂ = 0.7atm

The balanced chemical equation for the neutralization of HCl with is:

Given weight of = 5g

Moles of =

Volume of HCl solution =

Assuming the density of the solution is 1.0 g/mL

Mass of HCl solution = 50 g

Overall mass of the solution = 50 g + 5 g = 55 g

To find the heat of neutralization, we calculate:

Q = m C ΔT

where m equals the mass of the solution = 55 g

C represents the specific heat capacity of the solution = 4.184

ΔT signifies the temperature change = 6.8 K = (6.8 - 273) C = -266.2

The enthalpy of neutralization per mole of

=

Response:

Sulfate- SO4^2-

Sulfite- SO3^2-

Permanganate- MnO4

Carbonate- CO3^2

Clarification:

KEEP GOING WITH YOUR STUDIES!

Answer:

By reducing the height of the center of gravity of the object in relation to its center of buoyancy

Explanation:

In the field of hydrostatics, for a floating object, the state of equilibrium corresponds to either a peak or a trough in potential energy. Stability in equilibrium occurs when the potential energy is minimized. Achieving a lower position of the center of gravity of the floating object compared to its center of buoyancy creates a stable equilibrium arrangement.