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3 years ago

Write the balanced chemical Equation for the following when they are heated

Chemistry

1 answer:

ale4655 [162]3 years ago

4 0

Answer:

i. Lead nitrate:

2Pb (NO3)2 Δ= 2PbO+4NO2+O2

ii. Potassium chlorate:

2KClO3 → 2KCl + 3O2↑

iii. clacium carbonate:

CaCO3 + 2HCl -> CaCl2 + CO2 + H2O

iv. cupric carbonate :

CuCO3 → CuO + CO2↑

Hope this helped

All the best!!

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How many moles are there in 1.20 x 1025 particles of helium?

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Answer:

You would get 19.969 moles.

Explanation:

You would get 19.969 moles.

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Draw the structure of two alkenes that would yield 1-methylcyclohexanol when treated with Hg(OAc)2 in water, then NaBH4:Draw the

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Answer:

Two possible compounds are shown below- one with an exocyclic double bond and another one with an endocyclic double bond

Explanation:

Reaction of alkene with $Hg(OAc)_{2}$ gives a complex of mercurous ion.

Then water molecule attacks this complex through $S_{N}2$ type reaction at more substituted position.

$NaBH_{4}$ cleaves the resultant C-Hg bond and forms a C-H bond.

Two possible structures of an alkene is possible to yield 1-methylcyclohexanol which are shown below.

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Hydrogen gas, iodine vapor, hydrogen iodine are mixed in a flask and heated to 696°C. H2(g) + I2(g) ⇋ 2 HI(g) Kc = 52 at 696°C I

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Answer: The equilibrium concentration of hydrogen gas is 0.0269 M

Explanation:

The chemical reaction follows the equation:

$H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$

At t = 0 0.044M 0.044M 0.177M

At $t=t_{eq}$ (0.044-x)M (0.044-x)M (0.177+x)M

The expression for $K_c$ for the given reaction follows:

$K_c=\frac{[HI]^2}{[H_2]\times [I_2]}$

We are given:

$K_c=52$

Putting values in above equation, we get:

$52=\frac{(0.177+x)^2}{(0.044-x)^2}$

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Hence, the equilibrium concentration of hydrogen gas is (0.044-x) M =(0.044-0.0171) M= 0.0269 M

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3 years ago

Toluene (C6H5CH3 ), an organic compound often used as a solvent in paints, is mixed with a similar organic compound, benzene (C6

Sergeeva-Olga [200]

Explanation:

The given data is as follows.

Weight of solute = 75.8 g, Molecular weight of solute (toulene) = 92.13 g/mol, volume = 200 ml

Therefore, molarity of toulene is calculated as follows.

Molarity = $\frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}$

= 4.11 M

Hence, molarity of toulene is 4.11 M.

As molality is the number of moles of solute present in kg of solvent.

So, we will calculate the molality of toulene as follows.

Molality = $\frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}$

= 8.6 m

Hence, molality of given toulene solution is 8.6 m.

Now, calculate the number of moles of toulene as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{75.8 g}{92.13 g/mol}$

= 0.8227 mol

Now, no. of moles of benzene will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{95.6 g}{78.11 g/mol}$

= 1.2239 mol

Hence, the mole fraction of toulene is as follows.

Mole fraction = $\frac{\text{moles of toulene}}{\text{total moles}}$

= $\frac{0.8227 mol}{(0.8227 + 1.2239) mol}$

= 0.402

Hence, mole fraction of toulene is 0.402.

As density of given solution is 0.857 $g/cm^{3}$ so, we will calculate the mass of solution as follows.

Density = $\frac{mass}{volume}$

0.857 $g/cm^{3}$ = $\frac{mass}{200 ml}$ (As 1 $cm^{3}$ = 1 g)

mass = 171.4 g

Therefore, calculate the mass percent of toulene as follows.

Mass % = $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100$

= $\frac{75.8 g}{171.4 g} \times 100$

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Therefore, mass percent of toulene is 44.22%.

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3 years ago

Write the balanced chemical Equation for the following when they are heated ​

Write the balanced chemical Equation for the following when they are heated