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3 years ago

Explain why alkanes can be described as hydrocarbons.

Chemistry

1 answer:

worty [1.4K]3 years ago

5 0

Answer:

<h3>The International Union of Pure and Applied Chemistry (IUPAC) defines alkanes as "acyclic branched or unbranched hydrocarbons having the general formula CnH2n+2, and therefore consisting entirely of hydrogen atoms and saturated carbon atoms". ... The number of carbon atoms may be considered as the size of the alkane.</h3>

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Why is galvanized steel preferred for outdoor uses?

Marizza181 [45]

Galvanized steel is preferred for outdoor uses because it is ideal to prevent rotting/corrosion A steel will rot more quickly if it's exposed to a larger amount of oxygen and H2O , which will exist if we put it oudoor Coating the steel with additional zinc will slow down the process

7 0

3 years ago

A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from

sweet-ann [11.9K]

Answer:

$T_i~=163.1$ ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

Initial temperature of gold= Unknow

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the heat equation:

$Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT$

$Q_A_u=m_A_u*Cp_A_u*deltaT$

We can relate these equations if we take into account that all heat of gold is transfer to the water, so:

$m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT$

Now we can put the values into the equation:

$60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)$

Now we can solve for the initial temperature of gold, so:

$T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20$

$T_i~=163.1$ ºC

I hope it helps!

5 0

3 years ago

Picture attached, $3

Luba_88 [7]

Answer:

its -000.5

Explanation:

6 0

3 years ago

A diploid somatic ("body") cell has 2n = 20 chromosomes. At the end of mitosis, each daughter cell would have ______ chromosomes

kozerog [31]

Answer:

At the end of mitosis, 2n = 20

At the end of meiosis I, n = 10

At the end of meiosis II, n = 10

Explanation:

Mitosis is a type of cell division in which daughter cell produced are genetically identical to their mother cell. So, no. of chromosome does not change after mitosis.

So, at the end of mitosis, each daughter cell would have 20 chromosome.

Meiosis is a type of cell division in which mother cell produces two haploid cells ones with a single set of chromosomes.

Meiosis is a two step cell division, Meiosis I and Meiosis II.

In meiosis I, homologous pair separates, so no. of chromosomes becomes half.

In meiosis II, sister chromatids separates. So, the number of chromosomes remains same (i.e. Have same no. of chromosome as present in cell produced after meiosis I).

So, at the end of mitosis, each daughter cell would have 20 chromosome.

At the end of meiosis I, each daughter cell would have n = 10 chromosomes. At the end of meiosis II, each daughter cell would have n = 10 chromosomes.

7 0

3 years ago

Formula of sodium bicarbonate please help me with this

Stels [109]

Answer:

NaHCO₃

Explanation:

Sodium bicarbonate (baking soda) is a chemical compound with the formula NaHCO₃.

5 0

3 years ago