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Fynjy0 [20]
4 years ago
6

Patient added a 17-g measured dose of polyethylene glycol 3350 (MIRALAX) to 180 mL of water to use as a laxative. If the volume

of the resultant mixture was 195.6 mL, calculate the apparent density of polyethylene glycol 3350 and the specific gravity of the mixture
Chemistry
1 answer:
Sergio [31]4 years ago
8 0

Answer: Apparent density of ethylene glycol 3350 = 17g/15.6mL= 1.09g/mL

Specific gravity of mixture = 1.01

Explanation:

The specific gravity of an object or substance is the ratio of the density of that object or substance to that of a reference substance usually water.

<em>Specific gravity = density of substance/density of water</em>

The density of a substance is the ratio of the mass of that substance to is volume.

<em>Density = Mass/volume</em>

Density of water 1.00 g/mL,  

Apparent density of ethylene glycol 3350 = mass/volume

volume of ethylene glycol = volume of mixture - volume of water

volume of ethylene glycol = 195.6 - 180) = 15.6mL

Apparent density of ethylene glycol 3350 = 17g/15.6mL= 1.09g/mL

<em>mass of water =  density of water * volume of water</em>

mass of water = 1.00g/mL * 180mL = 180 g

Mass of solution = mass of substance + mass of water = (17 + 180)g = 197g

Density of mixture = mass of mixture / volume of mixture

Density of mixture = 197 g / 195.6 mL = 1.01 g/mL

specific gravity of mixture = density of mixture / density of water

specific gravity of mixture = 1.01g/mL / 1.00g/mL

specific gravity of mixture = 1.01

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Nutka1998 [239]

Answer:

Q1 = C * m * dT

Q2 = Qm * m

Qtotal = Q1 + Q2

Q1 - is amount of energy you need to apply to heat oxygen from the current temperature till you reach the melting temperature. Only if the oxygen is below to melting temperature.

C - is calorific capacity of oxygen -- better look at tables, it is a constant value

m - is the amount of oxygen, we will use moles because the other data shows moles, but could be grams, kg, etc.

dT - is the diference of temperatures between the current and the melting one. The melting temperature is constant and you can find it on tables, then (Tm - To)

Q2 is the amount of energy you have to add to melt oxygen once the oxygen has reached the melting temperature (Tm)

Qm is a constant value you could find on tables, depends on the mass of oxygen and is due to internal processes as changes in atomic distributions

If the oxygen is initially at melting temperature (melting point) you only need to know Q2, as dT = 0

I will do an example for you, but in future you should provide data of constants, it takes very long to find them in books or internet.

Data from tables

Tm =  54.36 K

C = 29.378 J/mol K this is at 25 C (or 298 K), is not really correct, you should look at its value at less than 54.36 K, but you can use it here.

Qm = 0.444 kJ/mol

Problem -- you have 44.33g of Oxygen -- Molecular weight of O2 is 32 g/mol

So you have 44.33/32 = 1.385 moles of oxygen

a) if oxygen is already at melting temperature: you only have to melt it

Qtotal = Q1 + Q2 = [0 (dT = 0) + Qm * m] = 0.444 * 1.385 = 0.615 kJ = 615 J

b) supposing an initial temperture of 50 K: now you have to heat oxygen till melting temperature and then melt it.

Q1 = C * m * dT = 29.378 * 1.385 * (54.36 - 50) = 177.442 J

Q2 = Qm * m = 615 J

Qtotal = 177.442 + 615 = 792.44 J

Explanation:

4 0
3 years ago
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Consider the reaction between iron (III) oxide, Fe2O3 and carbon monoxide, CO.
Lana71 [14]
<span>Fe2O3 + 3CO --> 2Fe + 3CO2
</span><span>
m(Fe2O3)=213 g
m(CO)=140 g
</span>_______________

<span>n(Fe2O3)=?
m(Fe)=?
n(Fe2O3)=?
n(CO)=?
n(CO2)=?
</span>
<span>n(Fe2O3)=m(Fe2O3) / M(Fe2O3)
n(Fe2O3)= 213 g / 159,7 gmol-1 = 1,33 mol
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<span>n(CO)= m(CO) / M(CO) n(CO)= 140 g / 28,01 gmol-1 = 4,99 mol</span>
3 0
3 years ago
What happens at the cathode in an electrolytic cell?<br><br> (Answer is reduction)
melisa1 [442]

Answer:

At the cathode in an electrolytic cell, ions in the surrounding solution are reduced into atoms, which precipitate or plate out on to the solid cathode. The anode is where oxidation takes place, and the cathode is where reduction takes place.

Explanation:

7 0
3 years ago
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I just need help on c-g. The whole question is there in case anyone needs it.
Elza [17]

Answer:

(c) 0.11; (d) -24.5 kJ·mol⁻¹; (e) See below; (f) Kc increases;

(g) No effect on ΔH

Step-by-step explanation:

(c) Kc at 125 °C  

In Part (a) one molecule of XY dissociated into one X and one Y.

       XY ⇌ X + Y

I:      10       0    0

C:     -1       +1   +1

E:      9        1     1

Kc = {[X][Y]}/[XY] = (1 × 1)/9 = ⅑ = 0.11

(d) ΔH

ΔH is a constant that is characteristic of the reaction.

ΔH = -24.5 kJ·mol⁻¹

(e) Effect of temperature on concentrations

If ΔH is negative, the reaction is exothermic.

Heat is a product of the reaction, so we can write the equation as

XY ⇌ X + Y + heat

If we lower the temperature, we are removing heat from the system.

<em>Le Châtelier's Principle</em> states that if you apply a stress to a system at equilibrium, it will respond by trying to relieve the stress.

We applied a stress by removing heat, so the system responds by producing more heat. The position of equilibrium moves to the right, and <em>more products will form</em>.

The diagram might look like the one below.

(f) Effect of temperature on Kc

Kc = [Products]/[Reactants]

We are increasing [Products] and decreasing [Reactants].

If you increase the numerator and decrease the denominator, you i<em>ncrease</em> the value of the quotient.

The value of the equilibrium constant increases when the temperature decreases.

(g) Effect of temperature on ΔH

Decreasing the temperature has no effect on ΔH, because the enthalpies of the reactants and products are properties of the substances themselves. They do not depend on the temperature.

7 0
3 years ago
If 23 atoms of hydrogen are present in the reactants, how many atoms of hydrogen will have to be present in the products?
madam [21]

Answer:

23

Explanation:

law of conservation of mass

quizlet

3 0
2 years ago
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