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3 years ago

What best describes the kinetic energy of two gas particles before and after a collision?

1 answer:

7 0

Kinetic energy is the energy possessed by the body when in motion. It is calculated by 1/2mv² where m is the mass of the body and v is the velocity.
Baring in mind that energy can not be created or destroyed but can only be transformed from one form to another (law of conservation of energy) then when two bodies collide the total kinetic energy of the two particles stays the same (kinetic energy is preserved). This is in the case of perfectly elastic collision, however, during inelastic collisions the kinetic energy is transformed to other forms of energy such as heat, sound and energy used in bond breakage.

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2. The combustion of heptane (C7H16) in oxygen spontaneously occurs. The products of the reaction are carbon dioxide and water v

masya89 [10]

The combustion of heptane (C7H16) in oxygen spontaneously occurs. The products of the reaction are carbon dioxide and water vapor. The balanced chemical equation is 7H16 + 11O2 → 7CO2 + 8H2O H = –2877.5 kJ.

8 0

3 years ago

A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 60.00 gram sample of the alcohol p

marin [14]

Answer:

$C_2H_6O$

Explanation:

The first step is the calculation of the moles of $H_2O$ and $CO_2$ , so:

$114.6~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=2.6~mol~of~CO_2$

$70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O$

Now, in 1 mol of CO2 we have 1 mol of C and in 1 mol of $H_2O$ we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to calculate the grams of C and O and then do the substraction form the initial amount, so:

$2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C$

$3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H$

$Total~grams=~31.25~+~7.82=39.08~g$

$grams~of~O=60.00~g-~39.08~g=20.92~g~of~O$

Now we can convert the grams of O to moles, so:

$20.92~g~of~O\frac{1~mol~O}{16~g~O}=1.30~mol~O$

The next step is to divide all the mol values by the smallest one:

$O=\frac{1.30~mol~O}{1.30~mol~O}=~1$

$C=\frac{2.6~mol~C}{1.30~mol~O}=~2$

$H=\frac{7.82~mol~H}{1.30~mol~O}=6$

Therefore the formula is $C_2H_6O$

6 0

3 years ago

Rank the compounds in each set in order of increasing acid strength.

Viktor [21]

Answer:

See explanation

Explanation:

For this question, we have to remember the effect of an atom with high electronegativity as "Br". If the "Br" atom is closer to the carboxylic acid group (COOH) we will have an inductive effect. Due to the electronegativity of Br, the electrons of the C-H bond would be to the Br, then this bond would be weaker and the compound will be more acid (because is easier to produce the hydronium ion $H^+$ ).

With this in mind, for A in the last compound, we have 2 Br atoms near to the acid carboxylic group, so, we will have a high inductive effect, then the C-H would be weaker and we will have more acidity. Then we will have the compound with only 1 Br atom and finally, the last compound would be the one without Br atoms.

In B, the difference between the molecules is the position of the "Br" atom in the molecule. If the Br atom is closer to the acid group we will have a higher inductive effect and more acidity.

See figure 1

I hope it helps!

5 0

4 years ago

Based on your answer in part B, arrange the five elements in order of their reactivity. Drag the most reactive element to the to

amm1812

Answer: Order of the elements from most reactive to least reactive is:

Mg > Zn > Fe > H > Cu

Explanation: Reactivity of the elements is the tendency of the elements to loose or gain electrons with ease.

As, all the given elements are metals, so the metals which will loose electrons easily will be more reactive. To study the reactivity of metals, we use reactivity series.

The metals which are more reactive are placed on the top of the reactivity series and which are less reactive are placed at the bottom.

Hence, the order of the elements from most reactive to least reactive is:

Mg > Zn > Fe > H > Cu

Reactivity series is attached below.

6 0

4 years ago

Iron and vanadium both have the BCC crystal structure, and V forms a substitutional solid solution for all concentrations at roo

mezya [45]

Answer:bbn

Explanation:

Nnnjjm

3 0

3 years ago