Cell differentiation starts from ------ day of conception. a)14th b) 23rd c) 20th d) 24th​

Answers 1

Answer:

Cell differentiation starts from 5th - 7th day of conception.

Question: Cell differentiation starts from 5th - 7th day of conception

Chapter Name: Cell Biology And Biotechnology

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The number of significant figures in 2345000 is * a) 7 b)6 c)5 d)4​

Answer:

seven is correct answer

Answer:

4

Explanation:

acc. to the rule

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Increasing marginal returns come to an end when __________./ वाढत्या किरकोळ परताव्याची समाप्ती होते जेव्हा __________. * 2 points AP intersects MP/ AP MP ला छेदतो TP begins to decline/ टीपी कमी होण्यास सुरुवात होते TP begins to rise at a diminishing rate/ टीपी कमी होण्याच्या दराने वाढू लागतो MP becomes negative/ खासदार नकारात्मक होतो​

Answer:

The law of diminishing marginal returns states that adding an additional factor of production results in smaller increases in output. After some optimal level of capacity utilization, the addition of any larger amounts of a factor of production will inevitably yield decreased per-unit incremental returns.

factor of production results

A. Find these words in the text. Then match them to the correct meaning. 1. bulge a. to capture or control something in order to use it 2. harness b. to swell; to rise up higher than normal 3. current c. a kind of wall built across a body of water 4. gravitational d. the movement of water in one direction 5. barrage e. relating to the attraction of objects to each other a​
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The heat of neutralization of hcl is 13.7 kcal and that of naoh and ch3cooh is 13.2 kcal. Heat of ionisation of ch3cooh is

Answer:

The heat released in the neutralisation of HCI and NaOH is 13.7 kcal/mol, the heat. released on neutralisation of NaOH with CH3COOH is 3.7 kcal/mol.

Answer:

The heat of ionization of CH₃COOH is 0.5kcal.

Explanation:

  • HCl is a strong acid and NaOH is a strong base. So, when NaOH reacts with CH₃COOH which is a weak acid, energy is only required to form CH₃COO⁻ +H⁺. And the reaction is given as,

[tex]CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O[/tex]     (1)

[tex]E=T-\Delta H[/tex]     (2)

Where,

E=heat of neutralization of the given reaction

T=total heat of neutralization of strong base and strong acid

ΔH=heat of ionization

From the question we have,

E=13.2kcal

T=13.7kcal

By substituting the value of E and T in equation (2) we get;

[tex]13.2kcal=13.7kcal-\Delta H[/tex]

[tex]\Delta H=13.7-13.2=0.5kcal[/tex]

Hence, the heat of ionization CH₃COOH is 0.5kcal.

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