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Tpy6a [65]
3 years ago
14

An atom that has 84 protons and 86 neutrons undergoes a reaction. At the end of the reaction, it has 82 protons and 84 neutrons.

What happened to the atom?
Chemistry
2 answers:
Ivenika [448]3 years ago
5 0

C- it emitted an alpha particle in a nuclear reaction (just took the edg test)

mr_godi [17]3 years ago
3 0
<span>During a nuclear reaction, an atom must emit an alpha particle, which consists of 2 protons (positively charged) and 2 neutrons (neutrally charged). Thus, the alpha particle has a charge of +2 (there are no electrons, which are negatively charged, to balance out the two protons). When an atom emits an alpha particle, it will lose this particle, thus losing 2 protons and 2 neutrons as well. Since the atom in the equation goes from 84 protons to 82, and 86 neutrons to 84, it lost 2 protons and 2 neutrons, indicating that it emitted an alpha particle.</span>
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Which of the following aqueous solutions has the lowest pH? Why?
jenyasd209 [6]
Answer:

HCI

Explanation:

The solution with the lowest pH is HCI with a pH value of 1. It is understood that it is more acidic because it is a strong acid.
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How many gallons in 1.3 liter for gas tank?
Veseljchak [2.6K]

0.34342367

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What ecosystem is being described in the passage?
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Determine the oxidation state for each of the elements below. The oxidation state of ... silver ... in ... silver oxide Ag2O ...
KatRina [158]

Answer:

The oxidation state of silver in \rm Ag_2O is +1.

The oxidation state of sulfur in \rm SO_2 is +4.

Explanation:

The oxidation states of atoms in a compound should add up to zero.

<h3>Ag₂O</h3>

There are two silver \rm Ag atoms and one oxygen \rm O atom in one formula unit of \rm Ag_2O. Therefore:

\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}.

The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is -2. Silver oxide \rm Ag_2O isn't an exception. Therefore:

\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}.

Solve this equation for the (average) oxidation state of \rm Ag:

\text{Oxidation state of $\rm Ag$} = 1.

<h3>SO₂</h3>

Similarly, because there are one sulfur \rm S atom and two oxygen \rm O atoms in each \rm SO_2 molecules:

\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}.

The oxidation state of \rm O in \rm SO_2 is also -2, not an exception, either.

Therefore:

\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}.

Solve this equation for the oxidation state of \rm S here:

\text{Oxidation state of $\rm S$} = 4.

8 0
4 years ago
how much current would be measured in a circuit if the light bulb has a resistance of 6 ohms and a voltage of 36 volts
Stels [109]

Answer:

The right response is "6 A". A further explanation is given below.

Explanation:

The given values are:

Resistance,

R = 6 ohms

Voltage,

V = 36 volts

As we know,

⇒  V=IR

then,

⇒  I=\frac{V}{R}

On substituting the values, we get

⇒     =\frac{36}{6}

⇒     =6 \ A

8 0
3 years ago
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