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deff fn [24]
2 years ago
9

Particles of a liquid

Chemistry
1 answer:
Tcecarenko [31]2 years ago
4 0

Answer:

A

Explanation:

B. That's a solid

C. Liquid is the only thing that can have viscosity

D. Not necessarily the case

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mass \: of \: CO _{2}  = (12 + (16 \times 2)) = 44 \: g \\ mass \: of \: oxygen \: in \: CO _{2} = (16 \times 2) = 32 \\ \% \: of \: oxygen =  \frac{32}{44}  \times 100\% \\  = 72.7\%

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At the melting point of a substance, temperature ______ as heat is being added and the substance is changing from a solid to a l
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Draw the condensed structural formulas for all the possible haloalkane isomers that have four carbon atoms and a bromine.
meriva

<u>Answer:</u> The isomers are shown in the image below.

<u>Explanation:</u>

Isomers are defined as the chemical compounds having the same number and kinds of atoms but arrangement are different.

For the alkane having four carbon atoms and 1 bromine atom, the IUPAC name of the haloalkane is bromobutane

There are 4 possible isomers for the given haloalkane compound:

  1. 1-bromobutane
  2. 2-bromobutane
  3. 1-bromo-2-methylpropane
  4. 2-bromo-2-methylpropane

The isomers of the given organic compound is shown in the image below.

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3 years ago
Aluminum metal reacts with bromine, a red-brown liquid with a noxious odor. The reaction is vigorous and produces aluminum bromi
GuDViN [60]

<u>Answer:</u> The mass of bromine reacted is 160.6 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}        .....(1)

Given mass of aluminium = 18.1 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium}=\frac{18.1g}{27g/mol}=0.670mol

The chemical equation for the reaction of aluminium and bromide follows:

2Al+3Br_2\rightarrow 2AlBr_3

By Stoichiometry of the reaction:

2 moles of aluminium reacts with 3 moles of bromine gas

So, 0.670 moles of aluminium will react with = \frac{3}{2}\times 0.670=1.005mol of bromine gas.

Now, calculating the mass of bromine gas, we use equation 1:

Moles of bromine gas = 1.005 moles

Molar mass of bromine gas = 159.81 g/mol

Putting values in equation 1, we get:

1.005mol=\frac{\text{Mass of bromine}}{159.81g/mol}\\\\\text{Mass of bromine}=(1.005mol\times 159.81g/mol)=160.6g

Hence, the mass of bromine reacted is 160.6 grams.

5 0
3 years ago
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