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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

balu736 [363]

Explanation:

Given -

An organic compound gives H₂ gas with Na
On treatment with alkaline iodine it gives yellow ppt.
On oxidation with CrO₃/H⁺ forms an aldehyde (C₂H₄O)

To Find -

Name the compound and write the reaction involved

Now,

Let A be the organic compound.

Then,

A + Na → + H₂↑
A + I₂ → CHI₃ (yellow ppt.)
A + CrO₃ + H⁺ → C₂H₄O

Now,

Here we see that compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives aldehyde.

Functional group of aldehyde = —CHO

And It forms only 2 Carbon aldehyde it means, It is Ethanal (CH₃CHO).

Compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives ethanal.

It means,

We know that 1° alcohol on oxidation gives aldehyde.

Here it gives 2 Carbon aldehyde.

It means,

Here 2 Carbon and 1° alcohol is used.

Now,

Its cleared that Compound A is Ethanol.

Reaction Involved -

CH₃CH₂OH + Na → CH₃CH₂O⁻Na⁺ + H₂↑
CH₃CH₂OH + I₂ + OH⁻ → CHI₃↓ + HCOO⁻ + HI + H₂O
CH₃CH₂OH + CrO₃ + H⁺ → CH₃CHO

6 0

3 years ago

After balancing the following reaction under acidic conditions, how many mole equivalents of water are required and on which sid

nordsb [41]

Answer:

d. 8 moles of H2O on the product side

Explanation:

Hello,

In this case, we need to balance the given redox reaction in acidic media as shown below:

$MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\$

Then, we add the half reactions:

$2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0$

Thereby, we can see d. 8 moles of H2O on the product side.

Best regards.

4 0

3 years ago

A hypothetical metal crystallizes with the face-centered cubic unit cell. The radius of the metal atom is 234 picometers and its

AleksAgata [21]

Answer:

$\delta=101.13 g/cm^3$

Explanation:

As can be seen in the Figure in a face-centered cubic unit cell you have:

Six halves of atoms
Eight 1/8 of atom (1 in each corner)

In total:

$n_{atom}=6*0.5+8*\frac{1}{8}=4 atoms$

Now, each side of the cell is 234 picometers (2.34e-8 cm) long

$V_{cell}=L^3=(2.34e^{-8} cm)^3$

$V_{cell}=1.28*10^{-23}cm^3$

Atoms per cm3:

$n=\frac{4 atoms}{1.28*10^{-23}cm^3}$

$n=3.12*10^{23} atoms/cm^3$

Expressing in mass:

$\delta=3.12*10^{23} atoms/cm^3* \frac{1 mol}{6.022*10^{23}}*195.08 g/mol$

$\delta=101.13 g/cm^3$

6 0

3 years ago

When a solution of lead(II) nitrate is mixed with a solution of sodium chromate, a yellow precipitate forms. (a) Enter the balan

Damm [24]

Answer:

a) Pb(NO₃)₂(aq) + Na₂Cr₂O₄(aq) ⇄ 2NaNO₃(aq) + Pb(Cr₂O₄)(s)

b) 67.6%

Explanation:

a) Nitrate is the ion NO₃⁻, and lead(II) forms the ion Pb⁺², so the compound lead(III) nitrate is Pb(NO₃)₂. (First, the cation, then the anion, with charges replaced without the signal).

Chromate is the ion Cr₂O₄⁻² and sodium forms the ion Na⁺, so the sodium chromate is Na₂Cr₂O₄. Both of them are in solutions, so they will be in an aqueous state.

In the reaction, the anions and cations will replace and will form: NaNO₃ and Pb(Cr₂O₄). The nitrates formed by metals from group 1, such as sodium, are soluble, so it will not forme a precipitated. So, the precipitated is PbCr₂O₄, and the balanced reaction is:

Pb(NO₃)₂(aq) + Na₂Cr₂O₄(aq) ⇄ 2NaNO₃(aq) + Pb(Cr₂O₄)(s)

b) The molar masses are: Pb(NO₃)₃ = 331,2 g/mol; Na₂Cr₂O₄ = 162 g/mol; Pb(Cr₂O₄) = 323,2 g/mol.

First, let's find what is the limiting reactant, doing the stoichiometry calculus for the reactants. Let's suppose that Na₂Cr₂O₄ is the limiting so:

1 mol of Pb(NO₃)₂ ------------------------------ 1 mol of Na₂Cr₂O₄

Transforming to mass (mass = moles * molar mass):

331,2 g of Pb(NO₃)₂ ------------------------- 162 g/mol of Na₂Cr₂O₄

x ------------------------- 12.38

By a simple direct three rule:

162x = 4100.256

x = 25.3 g of Pb(NO₃)₂

This is higher than what is put in the reaction, so Pb(NO₃)₂ is the limiting reactant, and Na₂Cr₂O₄ is in excess. So, let's do the stoichiometric calculus for the limiting reactant and the solid formed:

1 mol of Pb(NO₃)₂ ----------------------- 1 mol of Pb(Cr₂O₄)

Transforming to mass:

331.2 g of Pb(NO₃)₂ ------------------- 323.2 g of Pb(Cr₂O₄)

11.39 g ------------------- y

By a simple direct three rule:

331.2 y = 3681.248

y = 11.115 g

The yield is the mass formed divided by the stoichiometric result multiplied by 100%:

yield = (7.52/11.115)*100% = 67.6%

7 0

3 years ago

What part of the atom determines if it will combine or break apart from other substances

enyata [817]

The combination of an atom with other atom is bonding. Or break apart from other substance is breaking of interaction.

The bonding or interaction (inter-atomic) are due orbitals and electrons present in these orbitals

The atom consists of

a) a central nucleus which has proton and neutrons: they do not take part in chemical reactions

b) the part other than nucleus which has shells, sub shells and orbitals

These orbitals have electrons.

So the outer part of nucleus takes part in any combination with other substance

7 0

3 years ago