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Brums [2.3K]
3 years ago
8

How is cloning different from hybridization?

Chemistry
1 answer:
kvasek [131]3 years ago
6 0

1. Hybridization is a method of sexual reproduction whereas cloning is a method of asexual reproduction.

2. Hybrid animals are sterile, but cloned animals are fertile.

3. Hybrid organisms contain DNA from both male and female parents but cloned organisms contain DNA from only one type of parent.

4. Hybrid has superior character over its parents (improved hybrid vigor) but clones are 100% identical to their parent.

5. Hybridization gives only one hybrid progeny, whereas through cloning unlimited identical organisms can be produced.

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What is the key difference between a liquid and a gas?
s2008m [1.1K]

B. intermolecular attractions.

6 0
3 years ago
If a 1.0x10-5 L sample of agas at 2.0x106 atm isreleased until it is equal to0.275 atm, what would thenew volume of the gas be?
Greeley [361]

Answer:

Explanation:

Here, we want to get the new volume of the gas

What we need to know is the law that connects volume and pressure at constant temperature

The law that supports this is the Boyle's law

It states that the volume of a given mass of gas is inversely proportional to its pressure at constant temperature

Mathematically, we can have this as represented as:

\begin{gathered} P_1V_1=P_2V_2 \\ P_1\text{ = 2.0 }\times10^{6\text{ }}\text{ atm} \\ V_1\text{ = 1.0 }\times10^{-5}\text{ L} \\ P_2\text{ =0.275 atm} \\ V_2\text{ = ?} \end{gathered}

We can proceed mathematically to solve as follows;

undefined

5 0
1 year ago
Which has a greater density: an extremely small diamond used for industrial purpose such as grinding or a two-carat diamond take
ExtremeBDS [4]

The question is incomplete, the complete question is shown in the image attached to this answer

Answer:

See explanation

Explanation:

15) The both diamonds- the diamond used as a grinding tool and the two -carat diamond both have the same density because density is an intensive or intrinsic property of a substance. It does not depend on the mass of the substance present.

16) Area of magnesium = Length * width = 0.104 * 10^2 cm * 2.52 * 10^-1 cm  = 2.62 cm^2

Density = mass/area * thickness

Density * area * thickness = mass

thickness = mass/Density * area

thickness = 0.104 g/ 1.74g/cm3 * 2.62  cm^2

thickness = 0.0228 cm

3 0
3 years ago
If a gas is at a pressure of 46 mm Hg and temperature of 640 K, what would be the temperature if the pressure was raised to 760
Olegator [25]

Answer:

10573.9K

Explanation:

Using pressure law equation;

P1/T1 = P2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

P1 = 46mmHg

P2 = 760mmHg

T1 = 640K

T2 = ?

Using P1/T1 = P2/T2

46/640 = 760/T2

Cross multiply

640 × 760 = 46 × T2

486400 = 46T2

T2 = 486400 ÷ 46

T2 = 10573.9K

8 0
3 years ago
What is the heat of reaction (ΔH°rxn) for the combustion of acetone (C3H6O) given the following thermochemical equations? 1. 3 C
nydimaria [60]

Answer : The enthalpy of combustion of C_3H_6O will be -1775 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The combustion of C_3H_6O will be,

C_3H_6O(l)+4O_2(g)\rightarrow 3CO_2(g)+3H_2O(l)    \Delta H_{comb}=?

The intermediate balanced chemical reaction will be,

(1) 3C(s)+3H_2(g)+\frac{1}{2}O_2(g)\rightarrow C_3H_6O(l)     \Delta H_1=-285.0kJ

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-394.0kJ

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-286.0kJ

Now we are reversing the reaction 1, multiplying reaction 2 and 3 by 3 and then adding all the equations, we get :

(1) C_3H_6O(l)\rightarrow 3C(s)+3H_2(g)+\frac{1}{2}O_2(g)     \Delta H_1=285.0kJ

(2) 3C(s)+3O_2(g)\rightarrow 3CO_2(g)    \Delta H_2=3\times -394.0kJ=-1182.0kJ

(3) 3H_2(g)+\frac{3}{2}O_2(g)\rightarrow 3H_2O(l)    \Delta H_3=3\times -286.0kJ=-858.0kJ

The expression for enthalpy of combustion of C_3H_6O will be,

\Delta H_{comb}=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4

\Delta H_{comb}=(285.0)+(-1182.0)+(-858.0)

\Delta H_{comb}=-1755kJ

Therefore, the enthalpy of combustion of C_3H_6O will be -1775 kJ

3 0
4 years ago
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