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Ludmilka [50]
3 years ago
8

The total number of oxygen atoms indicated by formula Cu3(PO4)2?

Chemistry
1 answer:
ASHA 777 [7]3 years ago
6 0

Answer:

There's a total of 8 oxygen atoms in the formula Cu3(PO4)2

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3 years ago
You have a mass of 54 kg and are sitting on a stool. The normal force the stool applies on you is ...
Karo-lina-s [1.5K]

Answer:

529.2 N

Explanation:

As we have studied the first law of motion, which states that every action has some reaction, equal in magnitude but having an opposite direction.

The force that is acting on the student will be due to gravitational force, that is equal to his weight.

                                              F=mg: 54kg x 9.8m/s^2 =529.2 N

So the weight of student is exerting downwards towards the stool and land. The stool will also exert a force on the student that will be equal in magnitude but opposite in direction, then it will be 529.2 N.

This is because the student is sitting in a constant state and all the weight is exerted on the stool.

Note: This answer is very generic supposing that all the weight of the student is on stool. But, if we suppose that student's legs are on floor so it means the force of gravity acting on the stool has become less because student's mass on stool is less. So the answer would be a force somehow less than 529.2 N.  However, since the question asked normal force, it would be weight of student in general terms.

Hope it helps!

7 0
3 years ago
The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c
crimeas [40]

Answer : The rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 785.0K = ?

Ea = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 785.0K

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

\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]

K_2=1.45\times 10^{-2}s^{-1}

Therefore, the rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

7 0
3 years ago
Based on what you’ve determined regarding one mole of sand and one mole of water, which statements must be correct? Select all t
Igoryamba

Answer:

-The mole is appropriate only for counting things that are very small.

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Explanation:

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2 years ago
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Well assuming we have all of these, earth
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