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2 years ago

Can you guys help me on my science homework??

Chemistry

1 answer:

natulia [17]2 years ago

3 0

Answer:

1. B 2. D 3. Jon could test which animals like certain foods.

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you and your friends are at the park on a sunny day, and you are playing on the slide. As you sold down, the hot slide burns you

zzz [600]

Metal is a great conductor of heat while rope does not collect heat from the radiation from the sun as much or contain it while if metal is out in the sun all day it just keeps getting hotter and hotter and when you touch the rope it does not have much heat to transfer but the metal slide is hot enough to irritate your skin

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7 0

3 years ago

Which compounds can be made with the reaction benzene via electrophilic aromatic substitution using isopropyl chloride and AlCl3

Hitman42 [59]

Answer:

isopropyl benzene (cumene)

Explanation:

The reaction of isopropyl chloride and AlCl3 and benzene belongs to the class of organic reactions known as the Friedel Kraft alkylation.

The mechanism of the reaction involves the formation of a tetrahedral complex [AlCl4]^-.

The electrophile now is the isopropyl group which attacks the benzene to yield the product.

4 0

2 years ago

How many grams of nan3 are required to produce 19.0 ft3 of nitrogen gas, about the size of an automotive air bag, if the gas has

Papessa [141]

The balanced chemical reaction is given as:

$2NaN_{3}(s)\rightarrow 2Na(s)+3N_{2}(g)$

Now, convert $19.0 ft^{3}$ into litres.

$1 ft^{3} = 28.3168$

So, $19.0 ft^{3} = 19\times 28.3168 = 538.0192 L$

Density is equal to the ratio of mass to the volume.

$D=\frac{M}{V}$

where, M = mass and V= volume $(538.0192 L)$

Substitute the value of density and volume in formula to get the value of mass.

$1.25 g/L=\frac{M}{538.0192 L}$

$1.25 g/L\times 538.0192 L= M$

$Mass = 672.524 g$

Now, number of moles of $N_{2}$ gas= $\frac{672.524 g}{28.02 g/mol}$

= $24.00 moles$

According to the reaction, 2 moles of sodium azide gives 3 moles of nitrogen gas.

Now, in 24.00 moles of nitrogen gas produced from= $\frac{2 moles of sodium azide}{3 moles of nitrogen gas}\times 24.00 moles$ of nitrogen gas, moles of sodium azide.