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Ostrovityanka [42]

2 years ago

11

Which of these molecules and polyatomic ions cannot be adequately described using a single Lewis structure? Check all that apply

.CO3 2-COH2HCNO3NH4+OF2

1 answer:

11Alexandr11 [23.1K]2 years ago

3 0

<h3><u>Answer;</u></h3>

CO3^2- and O3

<h3><u>Explanation;</u></h3>

Ozone, O3 is a molecule composed of three oxygen atoms joined together. Two oxygen atoms joined together form the basic oxygen molecule O2.
A polyatomic ion is a charged particle which has two or more atoms held together by covalent (sharing of pairs of electrons) bonds.
Carbonate (CO3^2-) is an example of polyatomic ion. It can not be adequately described using Lewis structure.

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The number placed in front of a compound to balance a chemical reaction is called?

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2 years ago

Why Ammonia is nut dried by passing through concentrated H2SO4.

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2 years ago

Use the de Broglie's Wave Equation to find the wavelength of an electron moving at 7.3 × 106 m/s. Please show your work. Note: h

ad-work [718]

Answer:

$\lambda=9.96\times 10^{-11}\ m$

Explanation:

The expression for the deBroglie wavelength is:

$\lambda=\frac {h}{m\times v}$

Where,

$\lambda$ is the deBroglie wavelength

h is Planck's constant having value $6.62607\times 10^{-34}\ Js$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

v is the speed of electron.

Given that v = $7.3\times 10^6\ m/s$

Applying in the equation as:

$\lambda=\frac {h}{m\times v}$

$\lambda=\frac{6.62607\times 10^{-34}}{9.11\times 10^{-31}\times 7.3\times 10^6}\ m$

$\lambda=\frac{10^{-34}\times \:6.626}{10^{-25}\times \:66.503}\ m$

$\lambda=\frac{6.626}{10^9\times \:66.503}\ m$

$\lambda=9.96\times 10^{-11}\ m$

6 0

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

Murljashka [212]

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.

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

6 0

2 years ago