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-Dominant- [34]

3 years ago

15

Maple syrup, which comes from the sap of maple trees, contains water and natural sugars. It's a clear, brown liquid and the suga

rs can’t be separated by filtration. Under which category will you classify it?

1 answer:

scoray [572]3 years ago

7 0

The answer to that is a solution

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How many electron energy shells are occupied in an unstable atom of silicon

drek231 [11]

Answer:

4

Explanation:

it will share its valence electrons with other elements to acquire noble gas confriguration of the nearest inert element.

8 0

3 years ago

What is name of HOPBr2 ?

USPshnik [31]

HOPBr₂ = <span>dibromophosphinous acid

hope this helps!

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5 0

4 years ago

35.10 g of aluminum hydroxide is allowed to react with 53.94 g of sulfuric acid as follows:2Al(OH)3(s) + 3H2SO4(aq) -------->

ikadub [295]

Answer:

62.586 gram

Explanation:

moles of Al(OH)3 = mass / molar mass = 35.1 / (27+17x3) = 0.45 mol

moles of H2SO4 = mass / molar mass = 53.94 / (2+32+16x4) = 0.55 mol

H2SO4 is the limiting reagent (reacts completely)

⇒ moles of Al2(SO4)3 is worked out by moles of H2SO4

moles of Al2(SO4)3 = moles of H2SO4 / 3 = 0.183 mol

mass of Al2(SO4)3 = mole x molar mass = 0.183 x (27x2 + 96x3) = 62.586 gram

8 0

3 years ago

Formal charge of Cl in CHCl3

Vsevolod [243]

FC(C) = 4 - 0.5*8 - 0 = 0) FC(Cl) = 7 - 0.5*2 - 6 = 0

6 0

3 years ago

Two linear hydrocarbons, Hexane (C6H14) and Heptane (C7H16), form pretty much an ideal solution at any composition. A solution i

Anettt [7]

Answer:

$y_{C_6H_{14}}=0.92$

Explanation:

Hello,

At first, we compute liquid-phase molar fractions:

$n_{C_6H_{14}}=463.96 g*\frac{1mol}{86g} =5.3949molC_6H_{14}\\n_{C_7H_{16}}=667.71 g*\frac{1mol}{100g} =6.6771molC_7H_{16}\\x_{C_6H_{14}}=\frac{5.3949}{5.3949+6.6771} =0.447\\x_{C_7H_{16}}=1-x_{C_6H_{14}}=0.553$

Now, by means of the fugacity concept, for hexane, for instance, we have:

$f_{C_6H_{14}}^V=f_{C_6H_{14}}^L\\y_{C_6H_{14}}p_T=x_{C_6H_{14}}p_{C_6H_{14}}$

In this manner, at 25 °C the vapor pressure of hexane and heptane are 0.198946 atm and 0.013912 atm repectively, thus, the total pressure is:

$p_T=x_{C_6H_{14}}p_{C_6H_{14}}+x_{C_7H_{16}}p_{C_7H_{16}}\\p_T=0.447*0.198946 atm +0.553*0.013912 atm=0.096622atm$

Finally, from the hexane's fugacity equation, we find its mole fraction in the vapour as:

$y_{C_6H_{14}}=\frac{x_{C_6H_{14}}p_{C_6H_{14}}}{p_T}=\frac{0.447*0.198946 atm}{0.096622atm} \\y_{C_6H_{14}}=0.92$

Best regards.

3 0

3 years ago

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