All matter has mass and takes up space.<br> True<br> False

A sample of iron with a mass of 200g releases 9840 cal when it freezes at it’s freezing point. What is the molar heat of fusion

antoniya [11.8K]

Answer:-

2747.7 Cal mol -1

Explanation:-

Molar heat of Fusion is defined as the amount of heat necessary to melt (or freeze) 1 mole of a substance at its melting point.

Atomic mass of Iron = 55.845 g mol-1

Mass of Iron = 200 g

Number of moles of Iron = 200 g / (55.845 g mol-)

= 3.581 moles

Heat released = 9840 Cal

Molar heat of Fusion = Heat released / Number of moles

= 9840 Cal / 3.581 moles

= 2747.7 Cal mol -1

8 0

3 years ago

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What are the steps of natural selection?

blagie [28]

Answer:

The five steps involved in the process of natural selection are

Variation • Inheritance • Selection • Time • Adaptation

☆anvipatel77☆

•Expert•

Brainly Community Contributor

5 0

2 years ago

A particular sample of pure iron requires 0.612 kJ of energy to raise its temperature from 30.°C to 51°C. What must be the mass

mixas84 [53]

Answer:

m = 65.637 g

Explanation:

Heat = 0.612 kJ = 612 J ( Converting to J by multiplying by 1000)

Initial Temperature = 30.°C

Final Temperature = 51°C

Temperature change = Final Temperature - Initial Temperature = 51 - 30 = 21°C

Mass = ?

The relationship between these quantities is given by the equation;

H = mCΔT

where c = 0.444 J/g°C

Inserting the values in the equation;

612 = m * 0.444 * 21

m = 612 / (0.444 * 21)

m = 65.637 g

3 0

3 years ago

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

3 years ago

Which statement is true about molarity and percent by mass? (3 points)

lakkis [162]

Answer: The correct option is, They are different units of concentration.

Explanation: Molarity : It is defined as the number of moles of solute present in one liter of solution.

5 0

3 years ago

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All matter has mass and takes up space. True False