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3 years ago

Write out the rule for determining total magnification of a compound microscope

1 answer:

6 0

Total magnification of a compound microscope is found by multiplying the power of the objective times the power of the eyepiece.

For example, a 40X objective and 10X eyepiece multiply to 40x10 = 400X total magnification.

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What is the molecular formula of a compound with an empirical formula P2O5 and a gram-molecular mass of 284 grams? And why?

Lana71 [14]

I really don’t know but

Phosphorus pentoxide is a white solid which does not have any distinct odour. The chemical formula of this compound is P4O10. However, it is named after its empirical formula, which is P2O5. The molar mass of phosphorus pentoxide corresponds to 283.9 g/mol.

5 0

3 years ago

What happens when an electron in its lowest energy level or ground state when it is by absorbs energy?

tester [92]

Explanation:

atom changes from a ground state to an excited state by taking on energy from its surroundings in a process called absorption. The electron absorbs the energy and jumps to a higher energy level. In the reverse process, emission, the electron returns to the ground state by releasing the extra energy it absorbed

8 0

3 years ago

EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK

mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = $5.0 X 10^{-4} M$

$Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}$

$alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}$

= $2.02 X 10^{-13}$

When,

pH = -log[H+] = 10.15

[H+] = $7.08 X 10^{-11} M$

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = $7.4 X 10^{-7}$ ; Ka6 = $4.3 X 10^-11$

$alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}$

= $5.12 X 10^{-23}$

4 0

3 years ago

An object has a total mechanical energy of 150 J. At point A, the object has a kinetic energy

nika2105 [10]

Answer:

60 J

Explanation:

The law of conservation of energy states that energy is neither created nor destroyed, just converted into different forms. This means the total mechanical energy of the object at point A will be the same as the total mechanical energy at point B, and the question tells us the total of that mechanical energy is 150 J. Note we are assuming no energy is lost from the system as heat.

At point B, if the potential energy is 90 J, the remainder of the 150 J total must be kinetic energy. KE = 150 J - 90 J = 60 J.

8 0

3 years ago

A 2.5 g sample of french fries is placed in a calorimeter with 500.0 g of water at an initial temperature of 21 °C. After combus

SIZIF [17.4K]

Q=m°C<span>ΔT
=(500g) x (1 cal/g.</span>°C) x (48°C-21°C) = 13500 cal
13500 cal / 1000 = 13.5 kcal

<span>"What is the caloric value (kcal/g) of the french fries?"
13.5 kcal/ 2.5 g = 5.4 kcal/g</span>

8 0

3 years ago