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yaroslaw [1]
2 years ago
12

In a laboratory experiment, the freezing point of an aqueous solution of glucose is found to be -0.325°C, What is the molal conc

entration of this solution?
Chemistry
1 answer:
Hatshy [7]2 years ago
3 0

0.17 M is the is the molal concentration of this solution

Explanation:

Data given:

freezing point of glucose solution = -0.325 degree celsius

molal concentration of the solution =?

solution is of glucose=?

atomic mass of glucose = 180.01 grams/mole

freezing point of glucose = 146 degrees

freezing point of water = 0 degrees

Kf of glucose = 1.86 °C

ΔT = (freezing point of solvent) - (freezing point of solution)

ΔT = 0.325 degree celsius

molality =?

ΔT = Kfm

rearranging the equation:

m = \frac{0.325}{1.86}

m= 0.17 M

molal concentration of the glucose solution is 0.17 M

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b) Write a few paragraphs describing the chemical reaction and explaining the energy change in the reaction. Your document shoul
Lena [83]

Answer:

The energy change in a chemical reaction is due to the difference in the amounts of stored chemical energy between the products and the reactants. This stored chemical energy, or heat content, of the system is known as its enthalpy.

Explanation:

if u want to u can give me the crown btw have a good day

5 0
2 years ago
Read 2 more answers
6. Un volumen de 1.0 mL de agua de mar contiene casi 4 x 10-12 g de Au. El volumen total de agua en los océanos es de 1.5 x 1021
Aleks [24]

Answer:

The total amount of Au is $ 2.0\times10^{24}

Explanation:

Given that,

Mass of 1.0 ml of Au m=4\times10^{-2}\ g

Total volume of water in oceans V=1.5\times10^{21}\ L

We need to calculate the volume in ml

Using given volume

V=1.5\times10^{21}\times1000\ mL

V=1.5\times10^{24}\ mL

We need to calculate the total mass of Au  

Using given data

1\ ml\ volume = 4\times10^{-2}\ g

1.5\times10^{24}\ ml=4\times10^{-2}\times1.5\times10^{24}

So, The total mass of Au is 6\times10^{22}\ g

The mass will be in ounce,

Mass=0.035274\times6\times10^{22}

Mass=2.12\times10^{21}\ ounce

The total amount of the Au Will be

Total\ amount=2.12\times10^{21}\times948

Total\ amount=2.0\times10^{24}

Hence, The total amount of Au is $ 2.0\times10^{24}

3 0
2 years ago
Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. T
GenaCL600 [577]

Answer:

CN^- is a strong field ligand

Explanation:

The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).

Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.

Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.

5 0
2 years ago
the height of a column of mercury in barometer is 754.3mm. what us the atmospheric pressure in atm. in kPa.
Aleksandr-060686 [28]
1 atm = 760mmHg
754.3 mmHg / 760 mmHg * 1atm = 0.99 atm
760 mmHg = 101.3 KPa
754.3 mmHg/ 760mmHg *101.3 KPa = 100.54 KPa

Hope this helps!
8 0
2 years ago
Zinc and magnesium metal each reacts with hydrochloric acid to make chloride salts of the respective metals, and hydrogen gas. a
kirill115 [55]
M=11.20 g
m(H₂)=0.6854 g
M(H₂)=2.016 g/mol
M(Mg)=24.305 g/mol
M(Zn)=65.39 g/mol
w-?

m(Mg)=wm
m(Zn)=(1-w)m

Zn + 2HCl = ZnCl₂ + H₂
m₁(H₂)=M(H₂)m(Zn)/M(Zn)=M(H₂)(1-w)m/M(Zn)

Mg + 2HCl = MgCl₂ + H₂
m₂(H₂)=M(H₂)m(Mg)/M(Mg)=M(H₂)wm/M(Mg)

m(H₂)=m₁(H₂)+m₂(H₂)
m(H₂)=M(H₂)(1-w)m/M(Zn)+M(H₂)wm/M(Mg)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}

m(H₂)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}

(1-w)/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}

1/M(Zn)-w/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}

w(1/M(Mg)-1/M(Zn))=m(H₂)/{M(H₂)m}-1/M(Zn)

w=[m(H₂)/{M(H₂)m}-1/M(Zn)]/(1/M(Mg)-1/M(Zn))

w=0.583 (58.3%)
5 0
2 years ago
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