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2 years ago

Identify the atom whose nucleus is composed of 56 protons and 82 neutrons

1 answer:

3 0

An atom is the smallest particle of an element that can take part in a chemical reaction. Its structure has a nucleus and energy levels. The nucleus of an atom contain protons and neutrons whereas the energy levels contain electrons.
For a particle containing 56 protons (atomic number) and 82 neutrons has 56 electrons since the number of protons and neutrons in an uncharged atom are equal.
The atom in this case is must be an isotope of Barium since barium has an atomic number of 56 .

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Iron Choose one: A. is alloyed with carbon and chromium to make stainless steel. B. was the first of the common metals (iron, al

Lostsunrise [7]

Answer:

A. is alloyed with carbon and chromium to make stainless steel.

Explanation:

Steel is an alloy formed mainly of iron and carbon but some other metals like chromium are also added in little amounts.

Is steel, the percentage of iron and carbon together is about 90% and the rest metals fall in the 10% part.

Although the cost of steel is low, it has a very high tensile strength and that's why it is used in tools, ships, buildings, trains and in various types of infrastructures.

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2 years ago

Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec

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The overall reaction is given by:

$Br_{2}(g) + 2 NO(g) \rightarrow 2 NOBr(g)$

The fast step reaction is given as:

$NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})$

The slow step reaction is given as:

$NOBr_{2}(g) + NO(g) \rightarrow 2 NOBr(g)$ (slow step $k_{2}$ )

Now, the expression for the rate of reaction of fast reaction is:

$r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]$

The expression for the rate of reaction of slow reaction is:

$r_{2}=k_{2}[NOBr_{2}] [NO]$

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as $[NOBr_{2}]$ takes place in this reaction.

The expression of rate of formation is:

$\frac{d(NOBr)}{dt}=r_{2}$

= $k_{2}[NOBr_{2}][NO]$ (1)

Now, consider that the fast step is always is in equilibrium. Therefore, $r_{1}=0$

$k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]$

$[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]$

Substitute the value of $[NOBr_{2}]$ in equation (1), we get:

$\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]$

= $k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]$

= $\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]$

Thus, rate law of formation of $NOBr$ in terms of reactants is given by $\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]$ .

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3 years ago

What has 6 valence electrons and 16 protons and ends withthe letter "r"

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Answer:

Sulphur

Explanation:

Sulphur(S) is the Sixteenth element of the periodic table and it has an electronic configuration of 2,8,6. This means it has six electrons in its outermost shell( valence electrons).

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2 years ago

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3 years ago

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Brrunno [24]

the answer is A.) because the scientific method is used to answer problems.

This is the steps,

STEP 1: ask a question

STEP 2: state your hypothesis

STEP 3: conduct an experiment

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STEP 5: make a conclusion

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3 years ago

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