The question incomplete , the complete question is:
A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Answer:
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Explanation:
Moles of urea = 
Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity of the urea solution ;

Mass of solvent = m
Volume of solvent = V = 200.0 mL
Density of the urea = d = 0.95 g/mL


(1 g = 0.001 kg)
Molality of the urea solution ;


The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Answer:
The correct answer is - option C.
Explanation:
Given: the melting point of HCl is
-114.8 °C, which suggests that below this temperature HCl will be solid.
and, since the boiling point of HCl is - 85.1 °C. It is also suggested that above this temperature HCl will be gas, Therefore.
Solid -114.8 - Ordered arrangement
Liquid -85.1c - Less orderly arranged
Gaseous - Least orderly arranged
Thus, at —90 °C, HCl will be present 'in the liquid state, At — 1 °C, HCl will be present in the gaseous state and at -129 °C, HCl will be present in the solid-state. So, the molecules will be organized in a more orderly manner
.
Thus, the correct answer is - option C
Answer:
it is possible to remove 99.99% Cu2 by converting it to Cu(s)
Explanation:
So, from the question/problem above we are given the following ionic or REDOX equations of reactions;
Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V
Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V
In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:
Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.
Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.
Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.
Thus, ΔG° = - 92640.
This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.
Answer: 2.7 grams
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
Given: mass of sodium hydrogen carbonate = 3.4 g
mass of acetic acid = 10.9 g
Mass of reactants = mass of sodium hydrogen carbonate+ mass of acetic acid = 3.4 + 10.9= 14.3 g
Mass of reactants = Mass of products in reaction vessel + mass of carbon dioxide (as it escapes)
Mass of carbon dioxide = 14.3 - 11.6 =2.7 g
Thus the mass of carbon dioxide released during the reaction is 2.7 grams.