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Tpy6a [65]
3 years ago
5

The diagram below shows some subatomic particles.

Chemistry
2 answers:
marysya [2.9K]3 years ago
7 0

Answer is: It is either a proton or a neutron.

Quark is a type of elementary particle and a fundamental constituent of matter.

Quarks form composite hadrons (protons and neutrons). Protons and neutrons are in the nucleus of an atom.

Hadrons include baryons (protons and neutrons) and mesons.

There are six types of quarks: up, down, strange, charm, bottom, and top.

Alina [70]3 years ago
6 0

The correct option is this: IT IS EITHER A PROTON OR A NEUTRON.

Each atom of an element is made up of three sub particles, these are: electrons, protons and neutrons. The protons and the neutrons are located inside the nucleus of the atom while the electrons are located outside the nucleus, where they orbit round the nucleus. Since the sub particle X given in the question is located in the nucleus of the atom, then it is either a proton or a neutron.

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When N,N-Dimethylaniline is treated with bromine, ortho and para products are observed. However, when N,N-Dimethylaniline is tre
lilavasa [31]

Answer:

See explanation below

Explanation:

To get a better understanding watch the picture attached.

In the case of the reaction with Bromine, the -N(CH₃)₂ is a strong ring activator, therefore, it promotes a electrophilic aromatic sustitution, so, in the mechanism of reaction, the lone pair of the Nitrogen, will move to the ring by resonance and activate the ortho and para positions. That's why the bromine wil go to the ortho and para positions, mostly the para position, because the -N(CH₃)₂ cause a steric hindrance in the ortho position.

In the case of the reaction with HNO₃/H₂SO₄, the acid transform the -N(CH₃)₂ in a protonated form, the anilinium ion, which is a deactivating of the ring, and also a strong electron withdrawing, so, the electrophile will go to the meta position instead.

Hope this helps.

6 0
2 years ago
Volcano help! *I GIVE THANKS!*
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A volcanic <em>eruption</em> occurs when the <em>pressure </em> in a magma <em>chamber</em> becomes so great it is released like a valve. Magma is released through the volcano's <em>cone</em> in an eruption of <em>lava</em> rocks (bombs) and ash. A volcanic <em>cone</em> develops over centuries as flowing <em>lava</em> from the active volcano <em>cools </em>to form layers of rock. 
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Ahawk
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3 years ago
For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin
erastovalidia [21]

Answer : The initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1

k=7.5\times 10^{-3}M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1

\text{Rate}=3.4\times 10^{-3}Ms^{-1}

Therefore, the initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

8 0
3 years ago
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Alecsey [184]

Explanation:

Your chemical equation should look like this:

Li3PO4 + AlF3 --> 3LiF + AlPO4

This is the balanced equation for a double-displacement reaction

4 0
3 years ago
Please help find out code?
Nat2105 [25]
You might need to take more pictures so we can see all the equations clearly
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2 years ago
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