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3 years ago

Which is most likely the goal of selective breeding on a farm

2 answers:

s2008m [1.1K]3 years ago

5 0

The goal was to have an increased growth rate and an increased meat production.
For example, someone might selectively breed a sheep for more/softer wool.

Igoryamba3 years ago

3 0

To produce more meat

You might be interested in

Methane undergoes combustion. Which products form?

Aleksandr [31]

Methane is a carbon compound which undergoes combustion to release energy and form bi production which are Carbon dioxide ( CO2 ) and  Water ( H20 ).

the balanced chemical equation for the reaction is : -

CH4 + 2 O2 =》CO2 + 2 H2O

3 0

3 years ago

HELPPPPPP 30 POINTS Scientists are currently debating if a new epoch based on modern human activity should be added to the geolo

Margarita [4]

Answer:

Anthropocene

Explanation:

Anthropocene epoch is sort of unofficial epoch that scientists have used to describe basically human made impact on geological scale.

here's more info:

https://www.earthsystems.com/the-anthropocene/

3 0

2 years ago

How many moles of 0.225 M CaOH2 are present in 0.350 L of solution?

weeeeeb [17]

Answer : The number of moles of solute $Ca(OH)_2$ is, 0.0788 moles.

Explanation : Given,

Molarity = 0.225 M

Volume of solution = 0.350 L

Formula used:

$\text{Molarity}=\frac{\text{Moles of }Ca(OH)_2}{\text{Volume of solution (in L)}}$

Now put all the given values in this formula, we get:

$0.225M=\frac{\text{Moles of }Ca(OH)_2}{0.350L}$

$\text{Moles of }Ca(OH)_2=0.0788mol$

Therefore, the number of moles of solute $Ca(OH)_2$ is, 0.0788 moles.

7 0

3 years ago

How many molecules are in 18moles of CH.

-Dominant- [34]

Answer:

1.08 x 10²⁵molecules

Explanation:

From the mole concept we know that ;

1 mole of any substance contains 6.02 x 10²³ molecules

This number is the Avogadro's number.

So;

18 mole of CH will contain:

Number of molecules of CH = 18 x 6.02 x 10²³ = 1.08 x 10²⁵molecules

The number of molecules is therefore 1.08 x 10²⁵molecules

8 0

3 years ago

Heyy guys, so basically i need help with stoichiometric calculation I will give you 100 points just to answer all of these answe

jeka94

Answer:

3. The mass of ethanol required is approximately 0.522869 g

The mass of ethanoic acid required is approximately 0.68156 g

4. The mass of iron (III) oxide required is approximately 285.952.189.095 tonnes

5. The mass of silver nitrate required is approximately 14.53 grams

6. The mass of copper oxide that would be needed is approximately 31.86 grams

7. a. The mass of the precipitate, Zn(OH)₂ formed is approximately 49.712 grams

b. The mass of the precipitate, Al(OH)₃ formed is approximately 13 grams

c. The mass of the precipitate, Mg(OH)₂, formed is approximately 14.579925 grams

Explanation:

3. The 1 mole of ethanol and 1 mole of ethanoic acid combines to form 1 mole of ethyl ethanoate

The number of moles of ethyl ethanoate in 1 gram of ethyl ethanoate, n = 1 g/(88.11 g/mol) = 1/88.11 moles

∴ The number of moles of ethanol = 1/88.11 moles

The number of moles of ethanoic acid = 1/88.11 moles

The mass of ethanol = (46.07 g/mol) × 1/88.11 moles = 0.522869 g

The mass of ethanoic acid in the reaction = 60.052 g/mol × 1/88.11 moles ≈ 0.68156 g

4. 1 mole of iron(III) oxide reacts with 1 mole of CO₂ to produce 1 mole of iron

The number of moles in 100 tonnes of iron= 100000000/55.845 = 1790670.60614 moles

The mass of iron (III) oxide required = 159.69 × 1790670.60614 = 285952189.095 g ≈ 285.952.189.095 tonnes

5. The number of moles of NaCl in 5 grams of NaCl = 5 g/58.44 g/mol = 0.0855578371 moles

The mass of silver nitrate required, m = 169.87 g/mol × 0.0855578371 moles ≈ 14.53 grams

6. The number of moles of CuSO₄·5H₂O in 100 g of CuSO₄·5H₂O = 100 g/(249.69 g/mol) ≈ 0.4005 moles

The mass of copper oxide required, m = 79.545 g/mol × 0.4005 moles ≈ 31.86 grams

7. a. The number of moles of NaOH in the reaction = 20 g/(39.997 g/mol) ≈ 0.5 moles

2 moles of NaOH produces 1 mole of Zn(OH)₂

0.5 moles of NaOH will produce 0.5 mole of Zn(OH)₂

The mass of 0.5 mole of Zn(OH)₂ = 0.5 mole × 99.424 g/mol = 49.712 grams

The mass of the precipitate, Zn(OH)₂ formed = 49.712 grams

b. 6 moles of NaOH produces 2 moles Al(OH)₃

20 g, or 0.5 mole of NaOH will produce (1/6) mole of Al(OH)₃

The mass of the precipitate, Al(OH)₃ formed, m = 78 g/mol×(1/6) moles = 13 grams

c. 2 moles of NaOH produces 1 mole of Mg(OH)₂, therefore;

20 g or 0.5 moles of NaOH formed (1/4) mole of Mg(OH)₂

The mass of the precipitate, Mg(OH)₂, formed, m = 58.3197 g/mol × (1/4) moles = 14.579925 grams

3 0

3 years ago

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