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LekaFEV [45]
2 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]2 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

You might be interested in
How many grams of iron(II) chloride are needed to produce 44.3 g iron(II) phosphate in the presence of excess sodium phosphate?
zalisa [80]

Answer:

47.2 g

Explanation:

Let's consider the following double displacement reaction.

3 FeCl₂ + 2 Na₃PO₄ → Fe₃(PO₄)₂ + 6 NaCl

The molar mass of Fe₃(PO₄)₂ is 357.48 g/mol. The moles corresponding to 44.3 g are:

44.3 g × (1 mol / 357.48 g) = 0.124 mol

The molar ratio of Fe₃(PO₄)₂ to FeCl₂ is 1:3. The moles of FeCl₂ are:

3 × 0.124 mol = 0.372 mol

The molar mass of FeCl₂ is 126.75 g/mol. The mass of FeCl₂ is:

0.372 mol × (126.75 g/mol) = 47.2 g

5 0
2 years ago
......................................
Alinara [238K]

Answer:

Explanation:

yes leon is correct because matter is anything that takes space and also has mass in it.

thks.

5 0
3 years ago
Determine the pH of a KOH solution made by mixing 0.251 g KOH with enough water to make 1.0 × 10 2 mL of solution
vesna_86 [32]

Answer:

pH = 12.65

Explanation:

From the given information:

number of moles =mass in gram / molar mass

number of moles of KOH = mass of KOH / molar mass of KOH

number of moles of KOH =  0.251 g / 56.1 g/mol = 0.004474 mol

For solution :

number of moles = Concentration × volume

concetration = number of moles/ volume

concetration = 0.004474 mol / 0.100 L

concetration  = 0.04474 M

We know that 1 moles KOH result into 1 mole OH⁻ ions

Therefore,  Molarity  of OH⁻ = 0.04474 M

Now,

pOH = -log[OH⁻]

pOH = -log (0.04474) M

pOH = 1.35

Similarly,

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 1.35

pH = 12.65

6 0
2 years ago
Select all the correct answers.
Andrei [34K]

Scientists use scientific notation to communicate extremely small measurements. Hence, option A is correct.

<h3>What scientific notation?</h3>

Scientific notation is a way of writing very large or very small numbers.

Scientists use scientific notation to represent very small or very large numbers because this notation increases the  

  • accuracy of measured quantities.

  • convenience in using the numbers.

  • the number of significant figures.

  • precision of measurements.

Hence, option A is correct.

Learn more about the scientific notation here:

brainly.com/question/18073768

#SPJ1

4 0
1 year ago
One mole of a monatomic ideal gas is subjected to the following sequence of steps: a. Starting at 300 K and 10 atm, the gas expa
Verdich [7]

Answer:

a) Q = 0; W = 0; ΔU = 0; ΔH = 0; ΔS = 0.09 atm.L/K

b) Q = 1250 J; W = 0; ΔU = 1250 J; ΔH = 1250 J; ΔS = -0.0235 atm.L/K

c) Q = 3653.545 J; W = - 3653.545 J; ΔU = 0; ΔH = 0; ΔS = - 3653.545 J

d) Q = - 2080 J; W = 830 J; ΔU = - 1250 J; ΔH = - 2080 J; ΔS = - 5.984 J/K

Explanation:

a) If there is a vacuum, the work is zero, as it is a free expansion, the volume increases, the pressure decreases, the temperature is constant and the internal energy is constant.

∴ n = 1 mole

∴ PV = RTn....ideal gas

∴ P1 = 10 atm

∴ R = 0.082 atm.L/K.mol

∴ T = 300 K = T2

∴ V2 = 3*V1

⇒ W = 0.....expands freely into vacuum

⇒ ΔU = Q = 0....first law

⇒ ΔS = -  nR Ln(P2/P1).....ideal gas

∴ V1*P1/T1 = V2*P2/T2

∴ T1 = T2 = 300 K

⇒ P2 = V1*P1 / V2 = V1*P1 / 3V1 = 10 atm/3 = 3.33 atm

⇒ ΔS = - (1mol)*(0.082 atm.L/K.mol) Ln ( 3.33/10)

⇒ ΔS = 0.09 atm.L/K

∴ ΔH = ΔU + (P2V2 - P1V1) = 0 + 0 = 0

b) heated reversibly at constant volume:

⇒ W = 0 ...at constant volume

∴ T2 = 400 K; T1 = 300 K

∴ V1 = V2

⇒ Q = ΔU = CvΔT....first law

∴ Cv = 12.5 J/K.mol.....monoatomic ideal gas

∴ ΔT = 400 - 300 = 100 K

⇒ Q = ΔU = 12.5 J/mol.K * 100K = 1250 J/mol * 1 mol = 1250 J

∴ ΔH = ΔU + PΔV = ΔU + 0 = 1250 J

∴ ΔS = - nR Ln (P2/P1)

∴ P2/T2 = P1/T1...constant volume

∴ P1 = 3.33 atm

⇒ P2 = P1*T2 / T1 = (3.33 atm)*(400K) / (300K) = 4.44 atm

⇒ ΔS = - (1mol)*(0.082atm.L/K.mol) Ln (4.44/3.33)

⇒ ΔS = - 0.0235 atm.L/K

c) reversibly expanded at constant temperature:

∴ T1 = T2 = 400K

∴ V2 = 3*V1

∴ ΔU = 0...constant temperature

⇒ Q = - W....fisrt law

∴ W = - ∫ PdV..... reversibly expansion

∴ P = nRT/V... ideal gas

⇒ W = - nRT ∫ dV/V

⇒ W = - nRT Ln (V2/V1)

⇒ W = - (1mol)*(8.314 J/K.mol) Ln (3)

⇒ W = - 9.134 J/K *400K = - 3653.545 J

⇒ Q = - W = 3653.545 J

⇒ ΔH = ΔU + P1V1 - P2V2 = 0 + nRT1 - nRT2 = 0 + 0 = 0

∴ ΔS = - nR Ln(P2/P1)

∴ P1 = 4.44 atm

⇒ P2 = V1*P1*T2/ V2*T1 = V1*(4.44atm)*(400K) / (3.V1)*(400K)

⇒ P2 = 4.44atm/3 = 1.48 atm

⇒ ΔS = - (1mol)*(8.314 J/mol.K) Ln (1.48/4.44)

⇒ ΔS = -9.134J/K * 400K = - 3653.545 J

d) reversibly cooled at constant pressure:

∴ T2 = 300 K;  T1 = 400 K

∴ P2 = P1

⇒ Q = ΔH = CpΔT

∴ Cp = 20.8 J/K.mol

∴ ΔT = 300 - 400 = - 100 K

⇒ Q = ΔH = 20.8 J/mol.K * ( -100K) = - 2080 J/mol * 1mol = - 2080 J

⇒ ΔU = nCvΔT = (1mol)*(12.5 J/mol.K)*( - 100K) = -1250 J

⇒ W = ΔU - Q = ΔU - ΔH = -1250 J - ( - 2080 J ) = 830 J

∴ ΔS = ∫ δQ/T = ∫ nCpdT/T

⇒ ΔS = nCp Ln (T2/T1)

⇒ ΔS = (1mol)*(20.8 J/mol.K) Ln (300/400) = - 5.984 J/K

7 0
3 years ago
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