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egoroff_w [7]
3 years ago
7

Why do particles settle down in suspension

Chemistry
1 answer:
fomenos3 years ago
6 0
No nkomkooehruurururuu is
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what is the mole fraction of neon in a mixture that contains 0.628 g of helium, 11.491 g of neon, and 7.613 g of argon?​
VMariaS [17]

Answer:

0.2

Explanation:

Given parameters:

Mass of helium = 0.628g

Mass of neon = 11.491g

Mass of argon = 7.613g

Unknown:

Mole fraction of neon = ?

Solution:

The mole fraction of an element is the number of moles of that element to the total number of moles in the gas mixture.

We need to calculate the number of moles of each element first;

 Number of moles  = \frac{mass}{molar mass}

Molar mass of Helium = 4g/mol

Molar mass of Neon = 20g/mol

Molar mass of Argon = 40g/mol

Number of moles of He = \frac{0.628}{4} = 0.16moles

Number of moles of Ne = \frac{11.491}{20} = 0.58moles

Number of moles of Ar = \frac{7.613}{40} = 0.19moles

Total number of moles = 0.16moles + 0.58moles + 0.19moles = 0.93moles

Mole fraction Neon = \frac{0.19}{0.93}  = 0.2

4 0
3 years ago
How many electrons would be found in the Ion who’s symbol is I-
Vesnalui [34]

Answer:

54

Explanation:

Given symbol of the element:

                   I⁻

Number of electrons found in an ion with the symbol:

  This is a iodine ion:

         For an atom of iodine:

                   Electrons  = 53

                   Protons  = 53

                   Neutrons  = 74

An ion of iodine is one that has lost or gained electrons.

For this one, we have a negatively charged ion which implies that the number of electrons is 1 more than that of the protons.

  So, number of electrons  = 53 + 1  = 54

The number of electrons in this ion is 54

6 0
2 years ago
When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin
loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point

K_f = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality =\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9

8.2^0C=1\times K_f\times 1.9

K_f=4.32^0C/m

Now Depression in freezing point for sodium chloride is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=20.0^0C = Depression in freezing point

K_f = freezing point constant  

m= molality = \frac{136g\times 1000}{950g\times 58.5g/mol}=2.45

20.0^0C=i\times 4.32^0C\times 2.45

i=1.9

Thus vant hoff factor for sodium chloride in X is 1.9

3 0
3 years ago
How many types of biodiversity Are there
Mashcka [7]
3
Species
Genetic diversity
Ecosystem diversity
8 0
3 years ago
Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el
givi [52]

Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains 6.022\times 10^{23} number of particles.

We know that:

Charge on 1 electron = 1.6\times 10^{-19}C

Charge on 1 mole of electrons = 1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C

Cu^{2+}+2e^-\rightarrow Cu

2\times 96500=193000C is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

14\times 1000g=14000g of copper is deposited by =\frac{193000}{63.5}\times 14000=42551181 C

To calculate the time required, we use the equation:

I=\frac{q}{t}

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

40.0=\frac{42551181 C}{t}\\\\t=1063779sec

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, 1063779s\times \frac{1hr}{3600s}=295hr

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

3 0
3 years ago
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