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3 years ago

What do you meaning seachants? write with example.

Chemistry

1 answer:

Artemon [7]3 years ago

8 0

Answer:

answer

Explanation:

simile is a figure of speech that compares two different things in an interesting way. The object of a simile is to spark an interesting connection in a reader's or listener's mind. A simile is one of the most common forms of figurative language. Examples of similes can be found just about anywhere from poems to song lyrics and even in everyday conversations.

Similes and metaphors are often confused with one another. The main difference between a simile and metaphor is that a simile uses the words "like" or "as" to draw a comparison and a metaphor simply states the comparison without using "like" or "as".

An example of a simile is: She is as innocent as an angel. An example of a metaphor is: She is an angel. Do you see the difference? The simile makes a direct comparison, the metaphor's comparison is implied but not stated.

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The most common source of copper (cu) is the mineral chalcopyrite (cufes2). how many kilograms of chalcopyrite must be mined to

ser-zykov [4K]

The formula of chalcopyrite is CuFeS₂.
The molar mass of chalcopyrite - 183.5 g/mol
The mass of Cu in 1 mol of chalcopyrite is - 63.5 g
When 183.5 g of chalcopyrite is mined, the mass of Cu obtained - 63.5 g
So if 63.g of Cu is needed mass of chalcopyrite required - 183.5 g
Then to gain 370 g of pure Cu - mass of chalcopyrite needed -183.5/63.5 x 370
therefore mass of chalcopyrite to be mined - 1069.21 g

4 0

4 years ago

if the chromosome number in the sporophyte generation of a plant is 32 what is its number in the gametophyte generation?

xz_007 [3.2K]

Chromosomes have hereditary function that are capable of self-duplication and they have thread-like structures seen inside the nucleus of plant and animal. In plants, they go through a cycle called alteration of generation that has multicellular stages. One is sporophyte that is a diploid, and the other is gametophyte with a haploid.

<span>Moreover, spores are produce through meiosis so it go through a process of chromosome reduction in order to create a haploid spores. While, gametes undergo mitosis and the structure is already a haploid, which means that the number of chromosomes will not change in order to produce haploid. In this case, the plant have both spores and gametes and the similarities is that, they are both singles celled and a haploid. Therefore, the number of the chromosome in the gametophyte generation is also 32.</span>

3 0

3 years ago

If a chemical appears on the periodic table, it is a(n), ____________

Andre45 [30]

0 plus 5 negative and accumulated

7 0

3 years ago

A 0.1014 g sample of a purified compound containing C, H, and, O was burned in a combustion apparatus and produced 0.1486 g CO2

Alina [70]

Answer: The empirical formula for the given compound is $CH_2O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.1486g$

Mass of $H_2O=0.0609g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 0.1486 g of carbon dioxide, $\frac{12}{44}\times 0.1486=0.0405g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.0609 g of water, $\frac{2}{18}\times 0.0609=0.00677$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1014) - (0.0405 + 0.00677) = 0.054 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.0405g}{12g/mole}=0.003375moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.00677g}{1g/mole}=0.00677moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.054g}{16g/mole}=0.003375moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.

For Carbon = $\frac{0.003375}{0.003375}=1$

For Hydrogen = $\frac{0.00677}{0.003375}=2.00\times 2$

For Oxygen = $\frac{0.003375}{0.003375}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 2 : 1

The empirical formula for the given compound is $C_1H_2O_1=CH_2O$

Thus, the empirical formula for the given compound is $CH_2O$

8 0

3 years ago

How many grams of al(oh)3 (molar mass = 78.0 g/mol) can be produced from the reaction of 48.6 ml of .15 m koh with excess al2(so

Fudgin [204]

Taking into account the reaction stoichiometry, 0.18954 grams of Al(OH)₃ are formed from the reaction of 48.6 mL (0.0486 L) of 0.15 M KOH with excess Al₂(SO₄)₃.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Al₂(SO₄)₃ + 6 KOH → 2 Al(OH)₃ + 3 K₂SO₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Al₂(SO₄)₃: 1 mole
KOH: 6 moles
Al(OH)₃: 2 moles
K₂SO₄: 3 moles

The molar mass of the compounds is:

Al₂(SO₄)₃: 342 g/mole
KOH: 56.1 g/mole
Al(OH)₃: 78 g/mole
K₂SO₄: 174.2 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
KOH: 6 moles ×56.1 g/mole= 336.6 grams
Al(OH)₃: 2 moles ×78 g/mole= 156 grams
K₂SO₄: 3 moles ×174.2 g/mole= 522.6 grams

<h3>Definition of molarity</h3>

Molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume:

Molarity= number of moles÷ volume

<h3>Mass of Al(OH)₃ formed</h3>

In firts place, you know that 48.6 mL (0.0486 L) of 0.15 M KOH react with excess Al₂(SO₄)₃.

Replacing in the definition, you can calculate the amount of moles of KOH that react:

0.15 M= number of moles÷ 0.0486 L

Solving:

0.15 M× 0.0486 L= number of moles

<u><em>0.00729 moles= number of moles</em></u>

Then, 0.00729 moles of KOH react with excess Al₂(SO₄)₃. So, following rule of three can be applied: if by reaction stoichiometry 6 moles of KOH form 156 grams of Al(OH)₃, 0.00729 moles of KOH form how much mass of Na₂SO₄?

$mass of Al(OH)_{3} =\frac{0.00729 moles of KOHx156 grams of Al(OH)_{3}}{6 moles of KOH}$