Which is a description of objects found in abundance between Mars and Jupiter?

How many grams of a 14.0% (w/w) sugar solution contain 62.5 of sugar?

Delicious77 [7]

Answer:

$m_{solution}=446.4g$

Explanation:

Hello,

In this case, by-mass percent is computed in terms of the mass of the solute and the mass of the solvent as shown below:

$\% m=\frac{m_{solute}}{m_{solution}} *100\%$

Thus, solving for the mass of the solution, we obtain:

$\frac{x}{y} m_{solution}=\frac{m_{solute}}{\% m}*100\%\\ \\m_{solution}=\frac{62.5g}{14.0\%}*100\% \\\\m_{solution}=446.4g$

Regards.

3 0

3 years ago

A closed vessel system of volume 2.5 L contains a mixture of neon and fluorine. The total pressure is 3.32 atm at 0.0°C. When th

Oxana [17]

Answer: moles of Ne = 0.149 moles

moles of F₂ = 0.221 moles

Explanation:

The process occurs at costant volume.

Neon is a monoatomic gas, Cv =3/2R, Flourine is a diatomic gas, Cv = 5/2R

n = PV/RT ; where n is total = number of moles, P is total pressure = 3.32atm, V is volume = 2.5L, R is molar gas constant = 0.08206 L-atm/mol/K = 8.314 J/mol/K, T is temperature = 0.0°C = 273.15K

n(total) = (3.32 atm)(2.5 L)/(0.08206 L-atm/mol/K)(273.15) = 0.3703 mol

For one mole heated at constant volume,

Change in entropy, ∆S = ∫dq/T = ∫(Cv/T)dT

From T1 to T2, Cvln(T2/T2) = Cvln(288.15/273.15) = 0.05346•Cv

So, for 0.3703 moles,

∆S = (0.3703 mol)(0.05346)Cv = 0.345 J/K

⇒ Cv = 17.43 J/mol/K for the Ne/F₂ mixture.

For pure Ne, Cv = (3/2)R = 1.5 • 8.314 J/mol/K = 12.471 J/mol/K

For pure F₂, Cv = (5/2)R = 2.5 • 8.314 J/mol/K = 20.785 J/mol/K

If Y is the mole fraction of Ne, Y can be determined by setting the observed entropy change (∆S) to the weighted average of the entropy changes expected for the two different gases in the mixture:

17.43 J/mol/K = Y * 12.471 J/mol/K + (1 – Y) * 20.785 J/mol/K

⇒ 20.785 J/mol/K – 8.314 J/mol/K * Y = 17.43J/mol/K

Y = 0.403 ; 1 – Y = 0.597

moles of Ne = (0.403)(0.3703 mol) = 0.149 moles

moles of F₂ = (0.597)(0.3703 mol) = 0.221 moles

7 0

3 years ago

What is the driving force for the reaction of HCl(aq) and NaOH(aq)?

Andru [333]

HCl and NaOH react based on the following equation:
HCl + NaOH ..............> NaCl + H2O

This reaction is known as a double replacement reaction. In this types of reactions, the electrolyte (and the solvent) is water and the driving force is mainly the formation of a non-electrolyte produce.

Based on this, the driving force in the above neutralization reaction is the formation if the non-electrolyte salt (NaCl)

8 0

3 years ago

Enter your answer in the provided box. Calculate the volume of a 1.420 M NaOH solution required to titrate 40.50 mL of a 1.500 M

Nikolay [14]

Explanation:

Below is an attachment containing the solution.

7 0

3 years ago

What main type of forces must be overcome between br2 when liquid br2 dissolves into ethanol?

Snezhnost [94]

Dipole-induced dipole forces.

In contrast to C2H5OH, which is a very polar molecule, bromine is non-polar and has zero dipole moment. Charges are separated within polar molecules.

When a nonpolar Br2 molecule interacts with a polar C2H5OH molecule, one half of the bromine molecule acquires a charge and the other half acquires an opposing charge. A dipole is created when two adjacent bromine molecules have different charges. Bromine molecules must thus overcome the intermolecular forces—also known as dipole-induced dipole forces—that form in order to dissolve in ethanol.

Find more on bromine at : brainly.com/question/865727

#SPJ4

8 0

2 years ago