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lilavasa [31]
3 years ago
7

the bonds in a molecule of oxygen are blank blank bonds, while the bonds in a molecule of water are blank blank bonds

Chemistry
1 answer:
Fudgin [204]3 years ago
5 0
Non polar covalent and polar covalent are the only thing I can think of
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Please hep its my last question
atroni [7]

Answer:Noble gases:

 are highly reactive.

 react only with other gases.

 do not appear in the periodic table.

 are not very reactive with other elements.

Explanation:Noble gases:

 are highly reactive.

 react only with other gases.

 do not appear in the periodic table.

 are not very reactive with other elements.

7 0
3 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
3 years ago
Complete the following table. Tell if the molecule is polar or nonpolar, draw the Lewis dot structure for the molecule, tell wha
andrew11 [14]

The complete table is shown in figure

a) NH3 is polar as the bonds between N and H are polar. Due to asymmetry in   the molecule the molecule is polar

The shape of molecule is trigonal pyramidal while its electronic geometry is tetrahedral.

b) CO2: it is a non polar molecule with polar bonds. The molecule becomes non polar as the dipole moment cancel each other. [Dipole moment is a vector quantity]

The shape is linear.

5 0
3 years ago
Read 2 more answers
New account if u added my account antonio716 this my new account new friends
Rina8888 [55]

Answer:

Okay

Explanation:

6 0
3 years ago
Read 2 more answers
A photon with 2.3 eV of energy can eject an electron from potassium. What is the corresponding wavelength of this type of light?
Ira Lisetskai [31]

Answer:

\lambda=540.16\ nm

Explanation:

Given that:

The energy of the photon = 2.3 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.60 × 10⁻¹⁹ J

So, Energy = 2.3\times 1.60\times 10^{-19}\ J=3.68\times 10^{-19}\ J

Considering

Energy=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

\lambda is the wavelength of the light being bombarded

Thus,  

3.68\times 10^{-19}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

\frac{3.68}{10^{19}}=\frac{19.878}{10^{26}\lambda}

3.68\times \:10^{26}\lambda=1.9878\times 10^{20}

\lambda=5.40163\times 10^{-7}\ m=540.16\times 10^{-9}\ m

Also,  

1 m = 10⁻⁹ nm

So,  

\lambda=540.16\ nm

3 0
4 years ago
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