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zzz [600]
3 years ago
7

Can artificial selection aid in the evolution of an organism?

Chemistry
1 answer:
stepan [7]3 years ago
6 0
As stated on google it says “ Farmers and breeders allowed only the plants and animals with desirable characteristics to reproduce, causing the evolution of farm stock. This process is called artificial selection because people (instead of nature) select which organisms get to reproduce. ... This is evolution through artificial selection.”
You might be interested in
Calculate the molality of a solution formed by adding 6.30 g NH4CL to 15.7 g of water
babymother [125]

Answer:

Molality = 7.5 mol/kg

Explanation:

Given data:

Mass of NH₄Cl = 6.30 g

Mass of water = 15.7 g (15.7/1000 =0.016 kg)

Molality = ?

Solution:

Formula of molality:

Molality = Moles of solute / mass of solvent in gram

Now we will first calculate the number of moles of solute( NH₄Cl )

Number of moles = mass/ molar mass

Molar mass of  NH₄Cl = 53.491 g/mol

Number of moles = 6.30 g/  53.491 g/mol

Number of moles =  0.12 mol

Now we will calculate the molality.

Molality = Moles of solute / mass of solvent in gram

Molality =  0.12 mol / 0.016 kg

Molality = 7.5 m

or        (m=mol/kg)

Molality = 7.5 mol/kg

6 0
3 years ago
A can contains a gas with a volume of 86 mL at 30oC. What is the volume in the can if it is heated to 65oC?
Ber [7]

Answer:

New volume of gas = 95.93 ml (Approx)

Explanation:

Given:

Old volume of gas = 86 ml

Old temperature = 30°C = 30 + 273 = 303 K

New temperature = 65°C = 65 + 273 = 338 K

Find:

New volume of gas

Computation:

V1T2 = V2T1

(86)(338) = (V2)(303)

New volume of gas = 95.93 ml (Approx)

8 0
3 years ago
What pressure will be exerted by 25 g of CO2 at temperature of 25°C and a volume of .50 L?
Grace [21]

Answer:

P = 27.9 atm

Explanation:

Given data:

Mass of CO₂ = 25 g

Temperature = 25°C (25+273.15 K = 298.15 K)

Volume of gas = 0.50 L

Pressure of gas = ?

Solution:

Firs of all we will calculate the number of moles of gas,

Number of moles = mass/molar mass

Number of moles = 25 g/ 44 g/mol

Number of moles = 0.57 mol

Pressure of gas :

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

P × 0.50 L = 0.57 mol × 0.0821 atm.L/ mol.K  × 298.15 K

P = 13.95 atm.L/ 0.50 L

P = 27.9 atm

4 0
3 years ago
What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen?
jasenka [17]

Answer:

71.372 g or 0.7 moles

Explanation:

We are given;

  • Moles of Aluminium is 1.40 mol
  • Moles of Oxygen 1.35 mol

We are required to determine the theoretical yield of Aluminium oxide

The equation for the reaction between Aluminium and Oxygen is given by;

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.

Therefore;

1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen

1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium

Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.

4 moles of aluminium reacts to generate 2 moles aluminium oxide.

Therefore;

Mole ratio Al : Al₂O₃ is 4 : 2

Thus;

Moles of Al₂O₃ = Moles of Al × 0.5

                         = 1.4 moles × 0.5

                         = 0.7 moles

But; 1 mole of Al₂O₃ = 101.96 g/mol

Thus;

Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol

                                            = 71.372 g

3 0
3 years ago
1. Compare masses: a) 0,4mol CO₂ and 0,6mol H₂O ; b) 0,135mol H₂SO₄ and 0,5mol HCI.
bixtya [17]

Answer:

you can now deduct which one is greater or smaller and by how much.

Explanation:

no of moles= mass/molar mass

1ai) 0.4 = m/ ( 12 + (16*2)

m= 0.4* 44

m= 17.6g

ii) 0.6= m/( 2*1 + 16)

m= 0.6 *18

m= 10.8g

b) 0.135 = m/ ( 2*1 +32 + (16*4)

m= 0.135* 98

m= 13.23g

ii) 0.5= m/ (1+35.5)

m= 0.5*36.5

m= 18.25g

2. Avogadro's Number = 6.02×10²³

1 mol of any element= 6.02×10²³ particles

a) 0.1 mol of H20= (6.02×10²³) * 0.1

= 6.02×10²² molecules

ii) 0.3 mol of CO2= (6.02×10²³) * 0.3

= 1.806 × 10²³ molecules

Ans: 0.3 mol of CO2

bi) 0.25 mol of HCl= (6.02×10²³) * 0.25

= 1.505 × 10²³ molecules

bii) - find the no of moles first:

no of moles= mass/molar mass

n= 3.4g/ 34g →mr of H2S in g=2+32= 34g

n= 0.l mol

- use the Avogadro Number.

0.1 mol of H20= (6.02×10²³) * 0.1

= 6.02×10²² molecules.

biii) here you're given the density, use it to find the mass of acetic acid.

ρ = 1049 g/ml

ρ = m/v, where v=5 ml

1049 = m/ 5

m= ρ*v

m= 1049*5

m= 5245 g

• convert this into moles.

mr of CH3COOH= 12 + 3+ 12+ 16+ 16+ 1

= 60

mr in g = 60g

n= m/mr

n= 5245/ 60

n= 87. 41666...

n= 87.4 moles

•using Avogadro's Number:

87.4 moles of acetic acid=(6.02×10²³)*87.4

= 2.25148* 10²⁵

= 2.25 * 10²⁵ molecules

thus, the ans for this is 5 ml of acetic acid.

7 0
2 years ago
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