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2 years ago

Given the two compounds propane and pentanone, the one with the higher boiling point is

1 answer:

6 0

Pentanone, because it will boil at around 100 to 102 degree C, about the same as water, while propane boils at about -42 degree C, a temperature way below both pentanone and water's freezing point.

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Find [cu2+] in a solution saturated with cu4(oh)6(so4) if [oh − ] is fixed at 2.3 ✕ 10−6 m. note that cu4(oh)6(so4) gives 1 mol

ki77a [65]

$Cu_4(OH)_6(SO_4)$

You have OH- conc = 2.3 ✕ 10−6 m
From the formula, you can observe the ratio of Cu2+ to OH- is 4 : 6 = 2:3

So, for 2.3 ✕ 10−6 m OH-
[Cu2+] = $\frac{2}{3} \times 2.3 \times 10^{-6}$

$= 1.53 \times 10^{-6} 
$

6 0

3 years ago

A stack of 15 pennies is immersed into a 100 mL graduate initially containing 20.6 mL of water. The volume in the graduate cylin

Gekata [30.6K]

Answer: 5.4

Explanation: The Law of Impenetrability says that two objects can't occupy the same space at the same time; therefore, the pennies and the water can't occupy the same space at the same time. Without the pennies the water level read 20.6 mL, dropping in the pennies gives a level of 26.0.

5 0

2 years ago

What is the orbital diagram for fluorine?

pogonyaev

Explanation:

The orbital for fluorine is 1s 2s 2p. All of the boxes should be filled COMPLETELY until you can't no more. How do you know if you can't anymore? By the atomic number of the element. The atomic number for fluorine is 9. So, you keep adding arrows starting from the first box all the way to the last until you have reached the atomic number, in this case, 9.

Answer:

8 0

3 years ago

Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. A mixtur

yuradex [85]

Answer: The limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For magnesium:

Given mass of magnesium = 41.0 g

Molar mass of magnesium = 24 g/mol

Putting values in equation 1, we get:

$\text{Moles of magnesium}=\frac{41.0g}{24g/mol}=1.708mol$

For iron(III) chloride:

Given mass of iron(III) chloride = 175.0 g

Molar mass of iron(III) chloride = 162.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) chloride}=\frac{175g}{162.2g/mol}=1.708mol$

The chemical equation for the reaction of magnesium and iron(III) chloride follows:

$3Mg+2FeCl_3\rightarrow 3MgCl_2+2Fe$

By Stoichiometry of the reaction:

3 moles of magnesium reacts with 2 moles of iron(III) chloride

So, 1.708 moles of magnesium will react with = $\frac{2}{3}\times 1.708=1.114mol$ of iron(III) chloride

As, given amount of iron(III) chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, magnesium is considered as a limiting reagent because it limits the formation of product.

Moles of excess reactant left (iron(III) chloride) = [1.708 - 1.114] = 0.594 moles

Now, calculating the mass of iron(III) chloride from equation 1, we get:

Molar mass of iron(III) chloride = 162.2 g/mol

Moles of iron(III) chloride = 0.594 moles

Putting values in equation 1, we get:

$0.594mol=\frac{\text{Mass of iron(III) chloride}}{162.2g/mol}\\\\\text{Mass of iron(III) chloride}=(0.594mol\times 162.2g/mol)=96.35g$

Hence, the limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

6 0

3 years ago

For doping silicon with boron, silicon specimen was kept in gaseous atmosphere containing B2O3 that maintained the B concentrati

Kay [80]

Solution :

From Fick's law:

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=N_a$

Mass balance: Exits = Accumulation

$-N_A A = \frac{dm}{dt}$

$-N_A A = \frac{dVp}{dt}$

$-N_A A = \frac{dV}{dt}p$

$-N_A A = \frac{dhA}{dt}$

$-N_A A = \frac{dh}{dt} \times Ap$

From the last step, area cancels out and thus leaves :

$-N_A = \frac{dh}{dt} \times p$

So now we can substitute the $N_A$ by the Fick's law

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=\frac{dh}{dt} p$

Substituting the values we get

$=\frac{-4 \times 10^{-13}}{0.0001} \times (3 \times 10^{26} - C_{A2}) = \frac{dh}{dt} \times 2.46$

$=-4 \times 10^{-9} \times( 3 \times 10^{26} - C_{A2}) = 0.0001 \times 2.46$

$= -7800 \times 4 \times 10^{-9} \times (3 \times 10^{26}-C_{A2})=0.000246$ $=-(3 \times 10^{26}-C_{A2}) = 7.8846 \times 100 \times \frac{1}{69.62} \times 6.022 \times 10^{23}$

$C_{A2} = 6.85 \times 10^{28} \ \text{ boron atoms} /m^3$

3 0

3 years ago