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2 years ago

What is the main idea and Detail/Evidence of Light microscopes

Chemistry

2 answers:

Vesna [10]2 years ago

7 0

hi I I don’t know

Explanation:

Dimas [21]2 years ago

7 0

The advantage of light microscopes (and stereomicroscopes in particular) is that objects can be looked at with little or no preparation. This makes them very useful for looking at living things, such as flower parts, insects, earthworms and human skin.

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Suppose that the C=O group in a peptide bond can be regarded as isolated from the rest of the molecule. Given that the force con

Stella [2.4K]

Answer:

a) v = 1497.2 cm^-1

b) v = 1465 cm^-1

Explanation:

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2 years ago

You remove the pot of boiling hot water and set it on a cooler part of the stove. which mechanism is responsible for cooling the

kolbaska11 [484]

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2 years ago

What are the the basic cloud types?

shutvik [7]

Answer:

they are 10 but I'll be listing three

Explanation:

cirrus,cirrocumulus and cirrostratus

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2 years ago

Say I grab a bunch of aluminum foil (9.5g) and ball it up. Aluminum has a specific heat of .9J/g*C. How much would the aluminum’

Brrunno [24]

Answer:

ΔT = 0.78 °C

Explanation:

Given data:

Mass of Al = 9.5 g

Specific heat capacity of Al = 0.9 J/g.°C

Temperature change = ?

Heat added = 67 J

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

67 J = 9.5 g × 0.9 j/g.°C × ΔT

67 J = 85.5 j/°C × ΔT

ΔT = 67 J / 85.5 j/°C

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7 0

2 years ago

A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

Where;

$\frac{f_2}{f_1}$ is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356$

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0

2 years ago

What is the main idea and Detail/Evidence of Light microscopes​

What is the main idea and Detail/Evidence of Light microscopes