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melisa1 [442]
2 years ago
7

A chemistry teacher has a 560-cm length of magnesium ribbon. For a class experiment, each student requires an 8.5-cm length of t

he ribbon. How many students may receive the required length?
Chemistry
1 answer:
mariarad [96]2 years ago
6 0
Total length of magnesium ribbon = 560 cm
A student requires 8.5 cm

Number of students who will receive the required length = 560/8.5 = 65.882 = 65 students
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3 years ago
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Explain whether your breakfast was a compound, homogeneous mixture, or heterogenous mixture.
leonid [27]


what did you eat for breakfast?

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3 years ago
Tips for memorizing the first 20 elements of the periodic table?​
Anna [14]

Answer:

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hydrogen              helium                     lithium                     beryllium

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7 0
2 years ago
Read 2 more answers
For which of the following aqueous solutions would one expect to have the largest van’t Hoff factor (i)? a. 0.050 m NaCl b. 0.50
ira [324]

Answer:

The van't hoff factor of 0.500m K₂SO₄ will be highest.

Explanation:

Van't Hoff factor was introduced for better understanding of colligative property of a solution.

By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.

a) For NaCl the van't Hoff factor is 2

b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]

Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.

c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.

7 0
3 years ago
Acenapthalene has the empirical formula C6H5. A solution of 0.515 g of acenapthalene in 15.0 g CHCl3 boils at 62.5oC. The normal
german

Answer:

The molecular formula of an ascenapthalene is C_{12}H_{10}

Explanation:

\Delta T_b=K_b\times m

\Delta T_b=K_b\times \frac{\text{Mass of acenapthalene}}{\text{Molar mass of acenapthalene}\times \text{Mass of chloroform in Kg}}

where,

\Delta T_f =Elevation in boiling point = (62.5-61.7)^oC=0.8^oC

Mass of acenapthalene = 0.515 g

Mass of CHCl_3 = 15.0 g = 0.015 kg (1 kg = 1000 g)

K_b = boiling point constant = 3.63 °C/m

m = molality

Now put all the given values in this formula, we get

0.8^0C=3.67 ^oC/m\times \frac{0.515}{\text{Molar mass of acenapthalene}\times 0.015kg}

\text{Molar mass of acenapthalene}=155.7875 g/mol

Let the molecule formula of the Acenapthalene be C_{6n]H_{5n}

6n\times 12 g/mol+5n\times 1 g/mol=155.7875 g/mol

n = 2.0

The molecular formula of an ascenapthalene is C_{12}H_{10}

4 0
3 years ago
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