1.37x1020 uranium atoms

What is the relationship between layering of fluids and density?

Olin [163]

Density of a liquid determines how it will layer (heaviest to lightest). If the liquid is least dense it will float to the bottom. Layers will remain separated because each liquid is actually floating on top of the more dense liquid beneath it.

7 0

3 years ago

A 21.8 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter, according to the following reaction. If the temperature ris

Blababa [14]

Answer: The heat capacity of calorimeter is $15.66J/^oC$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of ethanol = 21.8 g

Molar mass of ethanol = 46.07 g/mol

Putting values in above equation, we get:

$\text{Moles of ethanol}=\frac{21.8g}{46.07g/mol}=0.473mol$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

$q$ = amount of heat released = ?

n = number of moles = 0.473 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -1235 kJ/mol = $-1235\times 10^3J/mol$ (Conversion factor: 1 kJ = 1000 J)

Putting values in above equation, we get:

$-1235\times 10^3J/mol=\frac{q}{0.473mol}\\\\q=(-1235\times 10^3J/mol\times 0.473mol)=-584.16\times 10^3J$

To calculate the heat capacity of calorimeter, we use the equation:

$q=c\Delta T$

where,

q = heat absorbed by the calorimeter = $584.16\times 10^3J$

c = heat capacity = ?

$\Delta T$ = change in temperature = $T_2-T_1=62.3^oC-25^oC=37.3^oC$

Putting values in above equation, we get:

$584.16\times 10^3J=c\times 37.3^oC\\\\c=\frac{584.16\times 10^3J}{37.3^oC}=15.66J/^oC$

Hence, the heat capacity of calorimeter is $15.66J/^oC$

4 0

4 years ago

Pls tell me how to solve this!

Anna35 [415]

Answer:

12 seconds

Explanation:

Time taken by 50cm³ of oxygen to diffuse from pinhole

= 1 minute = 60 seconds

⠀

$\textsf{ Rate of oxygen} \sf (O_2) = \frac{50}{60}$

⠀

Let time taken by 50cm³ of hydrogen to diffuse from pinhole = t seconds

⠀

$\textsf {Rate of hydrogen } \sf(H_2) = \frac{50}{t}$

⠀

According to the formula

$\sf \frac{Rate \: of \: hydrogen(H_2)}{Rate \: of \: oxygen(O_2) } = \sqrt{ \frac{Molar \: mass \: of \: O_2}{Molar \: mass \: of \: H_2} }$

⠀

$\large \sf \frac{50}{t} \div \frac{50}{60} = \sqrt{ \frac{\cancel{32}\small 16}{\cancel2} } \\ \\ \sf \large \frac{ \cancel{50}}{t} \times \frac{60}{ \cancel{50}} = \sqrt{16} \\ \\ \sf \large \frac{60}{t} = 4 \\ \\ \sf \large \frac{ \cancel{60} \: \small12}{ \cancel4} = t \\ \\ \large \underline{ \boxed{ \tt t = 12 \: seconds}}$