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tamaranim1 [39]
2 years ago
7

What direction would this plate move? (the plate is gray in color)

Chemistry
1 answer:
strojnjashka [21]2 years ago
7 0

Answer:

from south to north thats what it look like

Explanation:

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For the reaction shown, calculate how many moles of NO2 form when each amount of reactant completely reacts.
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Answer:-  2.4 mol NO_2 .

Solution:- It asks to calculate the moles of NO_2 formed when 1.2 moles of N_2O_5 are reacted.

There is 2:4 mol ratio between N_2O_5 and NO_2 . So, the moles of reactant are multiplied by the mol ratio to get the moles of NO_2 . The calculations are shown below:

1.2mol N_2O_5(\frac{4mol NO_2}{2mol N_2O_5})

= 2.4 mol NO_2

So, 2.4 moles of NO_2 are formed when 1.2 moles of  N_2O_5 were reacted.

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3 years ago
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What four factors account for the abundance of carbon compounds
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A carbon atom and a hydrogen atom form what type of bond in a molecule?
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A carbon atom and a hydrogen atom would form a covalent bond in a molecule. Specifically it would be a Nonpolar covalent bond.
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2 years ago
which statement is true about the part of the electromagnetic spectrum that is visible to the human eye?
IgorLugansk [536]
Hello @Lucysrv18,

How are you doing? In this case, we know that Electromagnetic waves that are visible to the human eye is neither made up of two electromagnetic waves nor three. Therefore B and C is incorrect.

These electromagnetic waves are divided and they are our final two options. For sure, i can say that the wavelengths in a visible electromagnetic wave are not divided into nine, but instead it is divided into 7. 

The answer to your question is A. 

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6 0
2 years ago
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Initially, a particular sample has a total mass of 200 grams and contains 128 x 1010 radioactive nuclei. These radioactive nucle
lubasha [3.4K]

Explanation:

Formula to calculate how many particles are left is as follows.

              N = N_{0} (\frac{1}{2})^{l}

where,     N_{0} = number of initial particles

                              l = number of half lives

As it is given that number of initial particles is 128 \times 10^{10} and number of half-lives is 3.

Hence, putting the given values into the above equation as follows.

               N = N_{0} (\frac{1}{2})^{l}

                    = 128 \times 10^{10}(\frac{1}{2})^{3}

                    = 16 \times 10^{10}

or,                    1.6 \times 10^{11}

Thus, we can conclude that 1.6 \times 10^{11} particles of radioactive nuclei remain in the given sample.

In five hours we've gone through 5 half lives so the answer is:

particles

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