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Nadusha1986 [10]
1 year ago
15

Tumu’s class was given an assignment to feature a scientist that contributed to the development of the cell theory. The class de

cided to use images to feature the scientist of their choice. Which image would Tumu most likely use in his assignment to feature Rudolf Virchow?
Chemistry
2 answers:
Agata [3.3K]1 year ago
6 0

Answer:

the way root cells reproduce to increase root length

Explanation:

Ad libitum [116K]1 year ago
4 0

Answer: the way root cells reproduce to increase root length

Explanation:

Rudolf Ludwig Carl Virchowas known as the founder of social medicine and also the father of modern pathology.

Virchow posited that all cells are gotten from already existing cells and he used this in his work towards cellular pathology, as it was made clear that diseases takes place at the cellular level. He posited that the cells that are malfunctioning cause diseases.

Based on the above analysis, the image that Tumu would most likely use in his assignment to feature Rudolf Virchow is the way root cells reproduce to increase root length.

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Calculate the amount, in moles, of PO43- present at equilibrium when excess Sr3(PO4)2 is added to 750. mL 1.2 M Sr(NO3)2(aq). As
Crank

Answer:

1.8 × 10⁻¹⁶ mol  

Explanation:

(a) Calculate the solubility of the Sr₃(PO₄)₂

Let s = the solubility of Sr₃(PO₄)₂.

The equation for the equilibrium is

Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹

                         1.2 + 3s          2s

K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} =  1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\

(b) Concentration of PO₄³⁻

[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹

(c) Moles of PO₄³⁻

Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol

7 0
2 years ago
A compound that is composed of carbon, hydrogen, and oxygen contains 70.6% C, 5.9% H, and 23.5% O by mass. The molecular weight
zhannawk [14.2K]

Answer: The molecular formula will be C_8H_8O_2

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{70.6g}{12g/mole}=5.9moles

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.9g}{1g/mole}=5.9moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.5g}{16g/mole}=1.5moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{5.9}{1.5}=4

For H = \frac{5.9}{1.5}=4

For O =\frac{1.5}{1.5}=1

The ratio of C : H: O= 4: 4:1

Hence the empirical formula is C_4H_4O

The empirical weight of C_4H_4O = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{136}{68}=2

The molecular formula will be=2\times C_4H_4O=C_8H_8O_2

4 0
2 years ago
State whether each of the following will be more soluble in water or hexane i. Butane ii. Ch3cooh iii. K2so4
ioda

Explanation:

Solubility is determined by the principle , "like dissolves like" .

i.e. , if a compound is polar then it will dissolve in a polar compound only , and

if a compound is non - polar then it will dissolve in a non - polar compound only .

Hence , from the question ,

Water is a polar molecule , and hence it will dissolve only the polar molecule , i.e. , from the given options the polar molecule is , iii. K₂SO₄

Hexane , is a non - polar molecules ,  hence it will dissolve only the non polar molecule , i.e. , from the given options the non polar molecule is i. Butane .

3 0
2 years ago
Which response includes all of the following processes that are accompanied by an increase in entropy?
dlinn [17]

Answer:

1 and 3.

Explanation:

The entropy measures the randomness of the system, as higher is it, as higher is the entropy. The randomness is associated with the movement and the arrangement of the molecules. Thus, if the molecules are moving faster and are more disorganized, the randomness is greater.

So, the entropy (S) of the phases increases by:

S solid < S liquid < S gases.

1. The substance is going from solid to gas, thus the entropy is increasing.

2. The substance is going from a disorganized way (the molecules of I are disorganized) to an organized way (the molecules join together to form I2), thus the entropy is decreasing.

3. The molecules go from an organized way (the atom are joined together) to a disorganized way, thus the entropy increases.

4. The ions are disorganized and react to form a more organized molecule, thus the entropy decreases.

7 0
2 years ago
Pentacarbonyliron(0) (fe(co)5) reacts with phosphorous trifluoride (pf3) and hydrogen, releasing carbon monoxide: fe(co)5+2pf3+h
vladimir1956 [14]

The limiting reactant can be determined by calculating the moles supplied / moles stoich ratio and the lowest is the limiting reactant.

Fe(CO)5 ratio = [6 g / 195.9 g/mol] / 1

Fe(CO)5 ratio = 0.0306

 

PF3 ratio = [4 g / 87.97 g/mol] / 2

PF3 ratio = 0.0227

 

H2 ratio = [4 g / 2 g/mol] / 1

H2 ratio = 2

 

<span>We can see that PF3 has the lowest ratio, so it is the limiting reactant.</span>

6 0
2 years ago
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