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2 years ago

Mention two substance that sublines

1 answer:

8 0

Familiar substances that sublime readily include iodine , dry ice , menthol, and camphor. Sublimation is occasionally used in the laboratory as a method for purification of solids, for example, with caffeine.

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Dimensional analysis, Plz hell I don’t understand and plz explain

Anastasy [175]

Explanation:

30 lb is 480 ounces

34 mi/second is 54.718 kilometre/ second

455 lb/ gal is 54521.024 grams / litre

50 cl is 500 millilitres

55nm is 5.5 × 10^-6 centimetre

7 0

3 years ago

Speed of an electron to orbit radius wire linear charge density is _________

ch4aika [34]

Answer:

A long uniformly charged wire has charge density λ=0.16μλ=0.16μC/m.

7 0

3 years ago

In the laboratory you dissolve 13.9 g of potassium phosphatein a volumetric flask and add water to a total volume of 250mL.

tensa zangetsu [6.8K]

Answer:

Molarity of the solution? 0,262 M.

Concentration of the potassium cation? 0,786 M.

Concentration of the phosphate anion? 0,262 M.

Explanation:

Potassium phosphate (K₃PO₄; 212,27 g/mol) dissolves in water thus:

K₃PO₄ → 3 K⁺ + PO₄³⁻ <em>(1)</em>

Molarity is an unit of chemical concentration given in moles of solute (K₃PO₄) per liters of solution.

There are 250 mL of solution≡0,25 L

The moles of K₃PO₄ are:

13,9 g of K₃PO₄ × $\frac{1mol}{212,27 g}$ = 0,0655 moles of K₃PO₄

The molarity of the solution is:

$\frac{0,0655 moles}{0,25L}$ = 0,262 M

In (1) you can see that 1 mole of K₃PO₄ produces 3 moles of potassium cation. The moles of potassium cation are:

0,0655 moles×3 = 0,1965 moles

The concentration is:

$\frac{0,1965 moles}{0,25L}$ = 0,786 M

The moles of K₃PO₄ are the same than moles of PO₄³⁻, thus, concentration of phosphate anion is the same than concentration of K₃PO₄. 0,262 M

I hope it helps!

4 0

3 years ago

Lactose, C12H22O11, is a naturally occurring sugar found in mammalian milk. A 0.335 M solution of lactose in water has a density

Dmitry_Shevchenko [17]

The molality of the solution is 0.00037 m.

<h3>What is concentration?</h3>

The term concentration refers to the amount of solute in a solution.

We have the following information;

Molarity = 0.335 M

Density = 1.0432 g/mL

Temperature = 20 o C

The molality of the solution is obtained from;

m = 0.335 M × 1.0432 g/mL/ 1000(1.0432 g/mL) - 0.335 M (342 g/mol)

m = 0.344/1043.2 - 114.57

m = 0.344/928.63

m = 0.00037 m

Learn more about molality of solution: brainly.com/question/4580605

5 0

3 years ago

Answer these please ASAP need help no idea how to do these

STALIN [3.7K]

Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass = g

Cl₂:

Number of moles = Mass / molar masa

Number of moles = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

8 mol = Mass / 2 g/mol

Mass = 16 g

P₄:

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles = 1.6 g /48 g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles = 54 g / 18 g/mol

Number of moles = 3 mol

CO₂

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

NH₃

Number of moles = Mass / molar masa

Number of moles = 8.5 g / 17 g/mol

Number of moles = 0.5 mol

CaCO₃

Number of moles = Mass / molar masa

Number of moles = 100 g / 100 g/mol

Number of moles = 1 mol

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO → 2Fe + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

Fe : F₂O₃

2 : 1

1.78 : 1/2×1.78 = 0.89 mol

Mass of F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂ → 2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

Al : AlI₃

2 : 2

0.05 : 0.05

I₂ : AlI₃

3 : 2

0.1 : 2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂ → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

Now we will compare the moles of oxygen with lead and lead oxide.

Pb : O₂

2 : 1

0.03 : 1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass = 0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂ → 2PbO

6 0

3 years ago

Mention two substance that sublines​

Mention two substance that sublines