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andre [41]
3 years ago
14

From his experiments ,j.j Thomson concludes what?

Chemistry
2 answers:
Vladimir [108]3 years ago
7 0

Answer:

Thomson made the following conclusions: The cathode ray is composed of negatively-charged particles. The particles must exist as part of the atom, since the mass of each particle is only ∼ 20001​start fraction, 1, divided by, 2000, end fraction the mass of a hydrogen atom.

Explanation:

Andrews [41]3 years ago
4 0

Answer:

J J Thomson's experiment confirms the existence of corpuscles (electrons), by concluding that: cathode rays are charges of negative electricity carried by particles of matter.

Explanation:

J J Thomson's experiment with cathode ray tubes helped him to discover the electron (which Dalton did not know about).

Cathode ray is composed of negatively-charged particles. And these particles must exist as part of the atom, since the mass of each particle is only ∼ 20001​start fraction, 1, divided by, 2000, end fraction the mass of a hydrogen atom. And also, that the particles that made up the gases were universal.

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Using noble gas notation write the electron configuration for the silicon atom
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Answer:

The closest noble gas to silicon is Neon (Ne)

Explanation:

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What does it mean when a reaction is spontaneous apex?
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A reaction that is spontaneous means that it can produce products without the supply of energy. The total energy from the reactants reached the activation energy that allow the reaction proceed. An example of this is the decay of a diamond into graphite occuring readily.
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4 years ago
a chemist wants to combine carbon and iron to form steel. what type of bond will this chemist create between these two elements
iVinArrow [24]
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5 0
3 years ago
How many grams of H2O will be formed when 32.0 g H2 is mixed with 84.0 g of O2 and allowed to react to form water
Zarrin [17]

Answer:

94.58 g of H_2O

Explanation:

For this question we have to start with the reaction:

H_2~+~O_2~->~H_2O

Now, we can balance the reaction:

2H_2~+~O_2~->~2H_2O

We have the amount of H_2  and the amount of O_2 . Therefore we have to find the limiting reactive, for this, we have to follow a few steps.

1) Find the moles of each reactive, using the molar mass of each compound (H_2~=~2~g/mol~~O_2=~32~g/mol ).

2) Divide by the coefficient of each compound in the balanced reaction ("2" for H_2 and "1" for O_2).

<u>Find the moles of each reactive</u>

32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}=15.87~mol~H_2

84.0~g~of~O_2\frac{1~mol~of~O_2}{32~g~of~O_2}=2.62~mol~of~O_2

<u>Divide by the coefficient</u>

<u />

\frac{15.87~mol~H_2}{2}=7.94

\frac{2.62~mol~of~O_2}{1}=2.62

The smallest values are for H_2, so hydrogen is the limiting reagent. Now, we can do the calculation for the amount of water:

32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}\frac{2~mol~H_2O}{2~mol~H_2}\frac{18~g~H_2O}{1~mol~H_2O}=94.58~g~H_2O

We have to remember that the molar ratio between H_2O and H_2 is 2:2 and the molar mass of H_2O is 18 g/mol.

6 0
3 years ago
CHEMISTRY HELP <br><br> I don't understand this....please help? thank you in advance
Igoryamba

60 i believe i haven't did science in a while sorry if im wrong though... ._.

6 0
3 years ago
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