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Anestetic [448]
3 years ago
7

Write the appropriate symbol for each of the following isotopes: (a) Z 74, A 186; (b) Z 80, A 201; (c) Z 34, A 76; (d) Z 94, A 2

39.
Chemistry
1 answer:
beks73 [17]3 years ago
6 0

Answer:  a)  _{74}^{186}\textrm{W}

b) _{80}^{201}\textrm{Hg}

c)  _{34}^{76}\textrm{Se}

d) _{94}^{239}\textrm{Pu}

Explanation:

General representation of an element is given as:_Z^A\textrm{X}

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

Mass number is defined as the sum of number of protons and neutrons that are present in an atom.

Mass number = Number of protons + Number of neutrons

In an atom, when neutrons or protons are lost or gains, it directly affects the mass number of an atom.

Atomic number is defined as the number of protons or number of electrons that are present in an atom.

It is characteristic of a particular element.

Atomic number = Number of electrons = Number of proton

a) Z 74, A 186: _{74}^{186}\textrm{W}

b) Z 80, A 201: _{80}^{201}\textrm{Hg}

c) Z 34, A 76: _{34}^{76}\textrm{Se}

d) Z 94, A 239.: _{94}^{239}\textrm{Pu}

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How many molecules in 3.4 moles of NH4NO3?
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There are  20.5 x 10^24 molecules are present in 3.4 moles of NH4NO3.

<h3>How many molecules in 3.4 moles of NH4NO3?</h3>

We know that one mole of a substance has  6.022 × 10²³ molecules so in 3.4 moles of NH4NO3, we have 20.5 x 10^24 molecules if we multiply the  6.022 × 10²³ with 3.4.

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2 years ago
150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac
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Answer:

0.175\; \rm mol \cdot L^{-1}.

Explanation:

Magnesium chloride and silver nitrate reacts at a 2:1 ratio:

\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s).

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to  \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}.

The precipitate silver chloride \rm AgCl is insoluble in water and barely ionizes. Hence, \rm AgCl\! isn't rewritten as ions.

Net ionic equation:

\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}.

Calculate the initial quantity of nitrate ions in the mixture.

\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}.

Since nitrate ions \rm {NO_3}^{-} do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol.

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be (1/2) of the concentration in the original solution.

\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}.

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