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Levart [38]
3 years ago
8

What happens to traits when mutations happen

Chemistry
1 answer:
Wittaler [7]3 years ago
7 0
Physical or mental traits change
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What is a insect and name one.
lidiya [134]

Answer:

a ladybug I hope this helps you

7 0
2 years ago
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Which law relates to the ideal gas law?
professor190 [17]

<u>Answer:</u> The law that related the ideal gas law is \frac{V_1}{n_1}=\frac{V_2}{n_2}

<u>Explanation:</u>

There are 4 laws of gases:

  • <u>Boyle's Law:</u> This law states that pressure is inversely proportional to the volume of the gas at constant temperature.  

Mathematically,

P_1V_1=P_2V_2

  • <u>Charles' Law:</u> This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

  • <u>Gay-Lussac Law:</u> This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

  • <u>Avogadro's Law:</u> This law states that volume is directly proportional to number of moles at constant temperature and pressure.

Mathematically,

\frac{V_1}{n_1}=\frac{V_2}{n_2}

Hence, the law that related the ideal gas law is \frac{V_1}{n_1}=\frac{V_2}{n_2}

8 0
3 years ago
Read 2 more answers
Metallic elements can be recovered from ores that are oxides, carbonates, halides, or sulfides. Give an example of each type.
vredina [299]

Metallic elements which can be recovered from ores that are oxides, carbonates, halides, or sulfides are iron, zinc, silver and lead respectively.

Ores which is deposited in earth's crust which contain minerals and metals. Metals can be obtained economically and sold commercially.

<h3>How metals obtained from sulphide or carbonate ore? </h3>

As we get to know that it is easy to obtain metals from their oxides. So, firstly ores which is found in the form of carbonates and sulphide are converted into their oxides by using the process of calcination and roasting.

The metals which is in the middle of the activity series are moderately reactive. These metals are found in the crust of the earth is mainly found as oxides, sulphides, or carbonates.

A metal which can occur in the form of sulphide ore is lead.

A metal which can occur in the form of oxide ore is iron.

A metal which can occur in the form of carbonate ore is zinc.

A metal which can occur in the form of halides ore is silver.

Thus, we concluded that the metallic elements which can be recovered from ores that are oxides, carbonates, halides, or sulfides are iron, zinc, silver and lead respectively.

learn more about ore:

brainly.com/question/10306443

#SPJ4

6 0
1 year ago
What observations can you make about this group of people? Name one inference you can make from your observations.​
Stells [14]

Answer:

no no no who are these some look good but are black what is this

7 0
3 years ago
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Twenty five grams of Iron 3 oxide react with an excess of carbon monoxide to form 15 g of Fe. Carbon dioxide is the other produc
densk [106]
<h3>Answer:</h3>

Theoretical mass = 17.42 g

Percent yield of Fe = 86.11%

<h3>Explanation:</h3>

The equation for the reaction between iron (iii) oxide and carbon monoxide is given by;

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

We are required to calculate the theoretical yield and the percentage yield of Iron.

Step 1: Moles of iron (iii) oxide

Moles are given by dividing the mass of the compound by the molar mass.

Molar mass of Iron(iii) oxide = 159.69 g/mol

Moles of Iron(III) oxide = 25 g ÷ 159.69 g/mol

                                     = 0.156 moles

Step 2: Moles of Iron produced

From the equation 1 mole of Iron(iii) oxide reacts to produce 2 moles of Fe.

Therefore, the mole ratio of Fe₂O₃ to Fe is 1 : 2.

Thus, moles of Fe = Moles of Fe₂O₃ × 2

                              = 0.156 moles × 2

                              = 0.312 moles

Step 3: Theoretical mass of iron produced

To calculate the mass of iron we multiply the number of moles of iron with the relative atomic mass.

Relative atomic mass = 55.845

Mass of iron = 0.312 moles × 55.845

                    = 17.42 g

Step 4: Percent yield of iron

% yield = (Actual mass ÷ Theoretical mass)×100

            = (15 g ÷ 17.42 g) × 100 %

            = 86.11%

7 0
3 years ago
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