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3 years ago

Draw the bond-line (not Lewis) structure for all the constitutionally isomeric compounds having the given molecular formula.

Chemistry

1 answer:

Westkost [7]3 years ago

3 0

Answer:

See explanation and image attached

Explanation:

The formula here is C4H10.

When we talk about constitutional isomers, constitutional isomers are compounds having the same molecular formula but different atom-to-atom connectivity.

The two main constitutional isomers of the formula C4H10 are butane and 2-methylpropane (iso-butane). The both structures are shown in the image attached.

Image credit: socratic

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How many grams are in 8.3 moles of CaCl2?

r-ruslan [8.4K]

Answer:

(8.3×40)+(8.3×71)

921.3grames

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3 years ago

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For the H, D and T tabulate three sub-atomic particles found in each of them

jonny [76]

Explanation:

The three sub atomic particles are;

Neutrons, Electrons and Protons

Hydrogen - H

Neutrons: 0

Electrons: 1

Protons: 1

Deuterium - D

Neutrons: 1

Electrons: 1

Protons: 1

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3 years ago

The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd

kenny6666 [7]

Answer:

The answer is " $= 0.078 \ kg \ H_2$ ".

Explanation:

calculating the moles in $CH_4 =\frac{PV}{RT}$

$=\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol$

Eqution:

$CH_4 +H_2O \to 3H_2+ CO \ (g)$

Calculating the amount of $H_2$ produced:

$= 12.9 \ mol CH_4 \times \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2$

So, the amount of dihydrogen produced = $0.078 \frac{kg}{s}$

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3 years ago

15. What volume of CCI, (d = 1.6 g/cc) contain

anastassius [24]

Answer:

$\boxed{\text{(3) 9.6 L}}$

Explanation:

1. Moles of CCl₄

$n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}$

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

$m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}$

4. Volume of CCl₄

$V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}$

4 0

3 years ago

When benzene (c6h6) reacts with bromine (br2) bromobenzene(c6h5br) is obtained: c6h6 + br2 → c6h5br + hbr what is the theoretica

saveliy_v [14]

Answer A) : We have to calculate the number of moles of Benzene involved in the reaction,

30 g / 78 moles of benzene = 0.384 moles

For bromine it will be the same process,

65 g / 159.8 moles = 0.406 moles

By observing the reaction given above we can say that the reaction ratio of bromine and benzene is 1 : 1

We need to find the mass of bromobenzene,

which should be, 6(12) + 5 (1)+ 79.90 = 156.9 g/mol

So, the mass of bromobenzene will be 156.9 g/mol X 0.3846 mol = 60.343 g

Hence the theoretical yield will be 60.34 g

Answer B) : To calculate the actual yield we have to divide it with theoretical yield.

(56.7g / 60.343 g ) X100% = 93.96 %

Here, we can say that we got 93.96 % of actual yield.

As we know it is impossible to get 100% yield in any reaction.

3 0

3 years ago