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Lunna [17]
3 years ago
10

I really need help who ever has the time and skills plzzz help me

Chemistry
1 answer:
Volgvan3 years ago
7 0
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Students are completing a table about a particular subatomic particle that helps make up an atom. The students have filled in on
Semmy [17]
Answer:
<span> Its location is in the nucleus, because the particle is a proton or a neutron.</span>
5 0
3 years ago
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What is the maximum mass of P2I4 that can be prepared from 8.80g of P4O6 and 12.37g of iodine according to the reaction:
Dvinal [7]
The balanced chemical reaction would be as follows:

<span>5P4O6 +8I2 ---> 4P2I4 +3P4O10

We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:

8.80 g P4O6 (1 mol / </span><span>219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / </span><span>253.809 g ) = 0.05 mol I2

Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:

0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (</span><span>569.57 g / 1 mol) = 14.24 g P2I4</span>
8 0
3 years ago
Read 2 more answers
. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r
Tom [10]

Answer :  The energy released by an electron in a mercury atom to produce a photon of this light must be, 4.56\times 10^{-19}J

Explanation : Given,

Wavelength = 435.8nm=435.8\times 10^{-9}m

conversion used : 1nm=10^{-9}m

Formula used :

E=h\times \nu

As, \nu=\frac{c}{\lambda}

So, E=h\times \frac{c}{\lambda}

where,

\nu = frequency

h = Planck's constant = 6.626\times 10^{-34}Js

\lambda = wavelength = 435.8\times 10^{-9}m

c = speed of light = 3\times 10^8m/s

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

E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}

E=4.56\times 10^{-19}J

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, 4.56\times 10^{-19}J

3 0
3 years ago
Compare la configuración realizada con la que aparece en la tabla periódica y confirma si esta en
mart [117]

Answer:

?

Explanation:

4 0
3 years ago
Determine the mass of the H2CO3 produced.
FrozenT [24]

Answer:

62.03 g/mol

Explanation:

7 0
3 years ago
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