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2 years ago

Q3. Why do you think there were so many heated debates about cell theory? Click or tap here to enter text.

1 answer:

8 0

Why were there so many heated debates around the development of the cell theory? There were no concrete information about the development of the cell, thus they did not have the evidence to come to conclusions. People would argue about what theory was correct. who named the cell?

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what does the number of the places after the decimal in a measurement tell you about the precision of the instrument that record

snow_lady [41]

The more numbers after the decimal point there are, the more precise the instrument which recorded it is. For example, if one instrument during seismic activity records that the magnitude of the earthquake was 2.3, and another instrument recorded that it was 2.3645, the second instrument would have shown to be more precise.

3 0

1 year ago

Now consider the reaction when 45.0 g NaOH have been added. What amount of NaOH is this, and what amount of FeCl3 can be consume

gayaneshka [121]

The correct answer is that 1.125 mol of NaOH is available, and 60.75 g of FeCl₃ can be consumed.

The mass of NaOH is 45 g

The molar mass of NaOH = 40 g/mol

The moles of NaOH = mass / molar mass

= 45 / 40

= 1.125

Thus, 1.125 mol NaOH is available

3 NaOH + FeCl₃ ⇒ Fe (OH)₃ + 3NaCl

3 mol of NaOH react with 1 mol of FeCl₃

1.125 moles of NaOH will react with x moles of FeCl₃

x = 1.125 / 3

x = 0.375 mol

0.375 mol FeCl₃ can take part in reaction

The molar mass of FeCl₃ is 162 g/mol

The mass of FeCl₃ = moles × mass

= 0.375 × 162

= 60.75 g

Thus, the amount of FeCl₃, which can be consumed is 60.75 g

6 0

2 years ago

Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.

Lesechka [4]

<u>Answer:</u> The enthalpy of the reaction for the production of $CH_4$ is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

For the given chemical reaction:

$C(s)+2H_2(g)\rightarrow CH_4(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]$

We are given:

$\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ$

Hence, the enthalpy of the reaction for the production of $CH_4$ is coming out to be -74.9 kJ

3 0

2 years ago

A protein subunit from an enzyme is part of a research study and needs to be characterized. A total of 0.145 g of this subunit w

Hitman42 [59]

Answer:

The molar mass of the protein is 12982.8 g/mol.

Explanation:

The osmptic pressure is given by:

π=MRT

Where,

M: is molarity of the solution

R: the ideal gas constant (0.0821 L·atm/mol·K)

T: the temperature in kelvins

Hence, we look for molarity:

$0.138 atm=M(0.0821\frac{l*atm}{mol*K} )(28+273K)$

$M=\frac{0.138atm}{(0.0821\frac{l*atm}{mol*K} )(301K)}$ = =5.584×10⁻³mol/l

As we have 2 ml of solution, we can get the moles quantity:

Moles of protein: 5.584×10⁻³ $\frac{mol}{l}\frac{1l}{1000ml}$ ×2ml=1.117×10⁻⁵mol

Finally, the moles quantity is the division between the mass of the protein and the molar mass of the protein, so:

Moles=Mass/Molar mass

Molar mass= Mass/Moles= $\frac{0.145g}{1.117*10^{-5}mol}$ =12982.8 g/mol

8 0

2 years ago

Gaseous cyclobutene undergoes a first-order reaction to form gaseous butadiene. At a particular temperature, the partial pressur

Gnoma [55]

Answer:

41.3 minutes

Explanation:

Since the reaction is a first order reaction, therefore, half life is independent of the initial concentration, or in this case, pressure.

$t_{1/2}= \frac{0.693}{K}$

So, fraction of original pressure = $\frac{1}{2}^2$

n here is number of half life

therefore, $\frac{1}{8}= \frac{1}{2}^3$

⇒ n= 3

it took 124 minutes to drop pressure to 1/8 of original value, half life = 124/3= 41.3 minutes.

8 0

2 years ago

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Q3. Why do you think there were so many heated debates about cell theory? Click or tap here to enter text.​

Q3. Why do you think there were so many heated debates about cell theory? Click or tap here to enter text.