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3 years ago

Convert 15oc to K (l’ll mark you as brainlister)

Chemistry

2 answers:

hammer [34]3 years ago

5 0

Answer:

15oC to K

oC +273

15+273

288K

Liono4ka [1.6K]3 years ago

4 0

15+273.15=288.15K.......

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An example of polytamic anoin would be a. Ca+2 b. NH4+ c. F d. PO4-3

ella [17]

$PO_4^{-3}$ and $NH^{4+}$ are polyatomic anions composed of two or more atoms.

<h3>What is a polyatomic anion?</h3>

A polyatomic ion is an ion composed of two or more atoms. Examples: The hydroxide cation ( $OH^-$ ) and the phosphate cation ( $PO_4^{3-}$ ) are both polyatomic ions.

$PO_4^{-3}$ and $NH^{4+}$ are the polyatomic anions because both contain more than 1 atom.

Hence, $PO_4{-3}$ and $NH^{4+}$ are polyatomic anions composed of two or more atoms.

Learn more about the polyatomic anion here:

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2 years ago

Does ccl4 has resonance structures

krek1111 [17]

Answer:

No, CCl₄ is 4 covalent C-Cl single bonds with a Tetrahedral geometry.

Explanation:l

For resonance structures to exist the molecule must have alternating single-double bonds. H₂C = CH - CH₃ <=> H₃C - CH = CH₂ resents a simple compound with a resonance structure system. This means that the π-bond electrons are distributed across all carbons in the molecular backbone. I would recommend internet searching for Danial Weeks 'Pushing Electrons' for a comprehensive review of molecular resonance structures. It is a brief, but easy to follow treatment of simple to complex structures containing resonance systems.

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8 0

2 years ago

18 g of argon occupy 750 ml at a particular temperature and pressure. How many grams of methane would occupy the same volume at

frozen [14]

Answer:

7.21 grams is the mass of methane

Explanation:

We may use the Ideal Gases Equation to solve this:

P. V = n. R. T

Let's determine the moles of Ar

18 g . 1 mol/ 39.9 g = 0.451 mol

In both situations, volume, temperature and pressure are the same so the moles of methane will also be the same as Argon's.

Let's convert the moles to mass of CH4.

0.451 mol . 16g/1mol = 7.21 grams

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3 years ago

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Aliun [14]

The correct answer is c

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3 years ago

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

$E=\frac{R_y}{n^2}$

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit = $E_{11}$

$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

$=E'=0.38\times 10^{-20} J$

$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

$\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m$

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

3 0

3 years ago