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LekaFEV [45]
3 years ago
13

In 1838, botanist Matthias Schleiden, determined that all plants are composed of cells. In 1839, anatomist Theodor Schwann propo

sed that all animals are composed of cells. In 1855, biologist Rudolph Virchow added to Schleiden’s and Schwann’s observations and proposed that all living things are composed of cells. Which statement is also part of Virchow’s cell theory?
Chemistry
1 answer:
Margarita [4]3 years ago
6 0

Answer:

This question is incomplete; the complete part is:

A) All cell have a cell wall.

B) All cell arise from pre-existing cells.

C) All cell are capable of photosynthesis.

D) All cell can develop into any other type of cell.

The answer is B

Explanation:

The commonly known universal theory proposed in 1838 took the contribution from three remarkable scientists namely: botanist Matthias Schleiden, anatomist Theodor Schwann and biologist Rudolph Virchow. According to the question, Mathias discovered that all plants are made of cells, Schwann determined that all animals are made of cells while Virchow determined that all living things are composed of cells.

However, in addition to Virchow's discovery, he also discovered and proposed that "All cell arise from pre-existing cells", which till date forms part of the three components of the cell theory. The three parts are:

- Cell is the fundamental and basic unit of all living things.

- All living things are made up of one or more cells

- All cells arise from pre-existing cells

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<h3><u>Answer;</u></h3>

3.64 moles

<h3><u>Explanation;</u></h3>

From the equation;

2Cu + O2 → 2CuO

Mole ratio of Cu : CuO = 2: 2 equivalent to 1: 1

Moles of CuO = 3.64 Moles

Therefore;

moles of Cu will be ;

3.64 × 1 = 3.64 moles

Moles of Cu = <u>3.64 moles </u>

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State the periodic law, and explain its relation to electron configuration. (Use Na and K in your explanation.)
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Answer:

Explanation:

The period law state that when elements are listed in order of their atomic numbers, the elements fall into recurring groups, so that there is a recurrence of similar properties at regular intervals.

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Na 11 -1s2 2s2 2p6 3s1

K 19 - 1s2 2s2 2p6 3s2 3p6 4s1

They share similar chemical and physical properties. Na and K are very reactive metals, they can loose/donate their outermost electron to non metals in other to attain stable octet state.

The form ionic compound when they react with non metals.

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What is a chemical off of the periodic table that is sticky
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20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator
Eduardwww [97]

Answer:

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

Bromocresol green, color change from pH = 4.0 to  5.6

Explanation:

The equation for the reaction is :

C_2H_5NH_2_(_a_q_)     +     H^+_(_a_q_)   ---      C_2H_5NH_{3(aq)}^+

concentration of C_2H_5NH_{2(aq) = 10%

10 g of C_2H_5NH_{2(aq) in 100 ml solution

molar mass = 45.08 g/mol

number of moles = 10 / 45.08

= 0.222 mol

Molarity of C_2H_5NH_2(aq) = 0.222 \times \frac{1000}{100}mL

= 2.22 M

number of moles of C_2H_5NH_{2(aq) in 20 mL can be determined as:

= 20 mL \times  2.22 M= 44*10^{-3} mole

Concentration of C_2H_5NH_2(aq) = \frac{44*10^{-3}*1000}{20}

= 2.22 M

Similarly, The pKa Value of C_2H_5NH_{2(aq) is given as 10.75

pKb value will be: 14 - pKa

= 14 - 10.75

= 3.25

the pH value at equivalence point is,

pH= \frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C]

pH = \frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22]

pH = 5.21

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

8 0
3 years ago
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