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3 years ago

Describe a proton, neutron, and electron. Where are these particles located inside an atom.

2 answers:

4 0

A proton is a positively charged particle found within the nucleus of an atom, a neutron carries no charge and is also found in the nucleus. An electron is a very small negatively charged particle found in the outer "shells" or orbitals of the atom.

lukranit [14]3 years ago

4 0

A proton has a positive charge, a neutron is neutral, and an electron has a negative charge. Protons and electrons are located inside an atom.

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Which area should have the lowest temperatures? A. 1 B.2 C.3 D.4

pochemuha

Answer:

A. 1

Explanation:

The higher you get in the atmosphere the cooler the temperature will be. The reason it's actually colder is because, as you go up in the atmosphere, the Earth's atmosphere feels less pressure the higher up you go. So as the gas in the atmosphere rises it feels less pressure, which makes it expand.

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7 0

3 years ago

The second-order rate constant for the dimerization of a protein (P) P + P → P2 is 6.2 × 10−3/M · s at 25°C. Part 1 out of 2 If

siniylev [52]

Answer:

$r=1.59\times 10^{-10}\ M/s$

Explanation:

According to the law of mass action:-

The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.

Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.

The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.

From the reaction given that:-

$P+P\rightarrown P_2$

The expression for the rate is:-

$r=k[P]^2$

Given that;- k= $6.2\times 10^{-3}$ /Ms

[P] = $1.6\times 10^{-4}$ M

Thus,

$r=6.2\times \:10^{-3}\times \:\left(1.6\times \:\:10^{-4}\right)^2\ M/s=1.59\times 10^{-10}\ M/s$

8 0

3 years ago

What is the standard entropy change for the reaction below? 2 CO(g) + 2 NO(g) → N2(g) + 2 CO2(g) S o (CO(g)) = 197.7 J/(mol·K) S

photoshop1234 [79]

Answer:

$\Delta _RS^o=-230J/(mol*K)$

Explanation:

Hello,

In this case, for the given chemical reaction:

$2 CO(g) + 2 NO(g) \rightarrow N_2(g) + 2 CO_2(g)$

The standard entropy change is computed by subtracting the products standard enthalpy and the reactants standard enthalpy considering each species stoichiometric coefficients:

$\Delta _RS^o=S ^o_{N2(g)}+2S ^o_{CO2(g)}-2S ^o_{CO(g)}-2S ^o_{NO(g)}\\\\\Delta _RS^o=191.6J/(mol*K)+2*213.8J/(mol*K) -2*213.8J/(mol*K) -2*210.8J/(mol*K)\\\\\Delta _RS^o=-230J/(mol*K)$