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AVprozaik [17]
3 years ago
9

Number of H in 3(NH4)2Cro4

Chemistry
1 answer:
zlopas [31]3 years ago
7 0

The number of H atoms in 3(NH₄)₂CrO₄ = 24

<h3>Further explanation  </h3>

The empirical formula is the smallest comparison of atoms of compound forming elements.  

A molecular formula is a formula that shows the number of atomic elements that make up a compound.  

(empirical formula) n = molecular formula  

Subscripts in the chemical formula indicate the number of atoms

The compound of 3(NH₄)₂CrO₄ ( 3 molecules of (NH₄)₂CrO₄ ) :

Number of H :

\tt 4\times 2(subscript)\times 3(coefficient,number~of~molecules)=24~atoms

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Which of the following item(s) explain the differences between the Ka values. Choose one or more: A. The oxidation state for oxy
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Answer:

<em>C. The electron-withdrawing fluorine atoms pull electron density from the oxygen in trifluoroacetate. The negative charge is more stabilized in trifluoroacetate by this effect.</em>

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Explanation:

<em>The structures of trifluoroacetate and acetic acid are both shown in the image attached.</em>

<em>The trifluoroacetate anion (CF3CO2-), just like the acetate anion has in the middle, two oxygen atoms.</em>

<em>However, in the trifluoroacetate anion, there are also three electronegative fluorine atoms attached to the nearby carbon atom attached to the carbonyl, and these pull some electron density through the sigma bonding network away from the oxygen atoms, thereby spreading out the negative charge further. This effect, called the "inductive effect" stabilizes the anion formed,the trifouoroacetate anion is thus more stabilized than the acetate anion.</em>

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10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of
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First we have to calculate the new or final volume of carbon dioxide gas.

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The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

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P_1 = initial pressure of gas = 10 kPa

P_2 = final pressure of gas = 15 kPa

V_1 = initial volume of gas = 10m^3

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 50^oC=273+50=323K

T_2 = final temperature of gas = 75^oC=273+75=348K

Now put all the given values in the above equation, we get:

\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}

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Now we have to calculate the new density of carbon dioxide gas.

PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}

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\rho_2=\frac{P_2M}{RT_2}

where,

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\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}

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