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3 years ago

How could measuring the melting point of a solid help decide whether it was a mixture or a compound?

Chemistry

1 answer:

mars1129 [50]3 years ago

8 0

Answer: It can't.

Explanation:

In most cases, the melting point alone will not enable you to identify a compound. Millions of solid organic compounds, and their melting points, are known. Perhaps 10,000 of these will have the same melting point as your unknown compound.

Hope this helps!

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Scoring Scheme: 3-3-2-1 Part II. You considered the properties of two acid-base indicators, phenolphthalein and methyl orange. M

kotykmax [81]

Answer:

Explanation:

Phenolphthalein is a protonated indicator and methyl orange is a basic indicator having hydroxyl ionisable part .

Phenolphthalein can be represented by the following formula

HPh which ionizes in water as follows

HPh + H₂O ⇄ H₃O⁺ + Ph⁻

( colourless ) ( pink )

In acidic solution it is in the form of protonated Hph form which is colourless

In basic medium , it ionises to give H₃O⁺ and unprotonated Ph⁻ whose colour is pink .

7 0

3 years ago

What information does the formula for an ionic compound provide?

Korolek [52]

It provides us with the knowledge of what the cation and anion of the compound are, as well as how many atoms of each are present.

7 0

2 years ago

I AM GIVING BRAINLIEST PLEASEE HELPPPP I NEED HELPPPPPP PLEAEEEEEE

True [87]

Answer:

0.479 M or mol/L

Explanation:

So Molarity is moles/litres of solution...often written as M=mol/L

So here we are given grams of BaCl2 which we have to convert to moles. To convert to moles of BaCl2 we have to divide 63.2 g BaCl2 by molar mass of BaCl2 which is 208.23 g/mol so you get 63.2/208.23 = 0.3035 moles of BaCl2

Second step is converting the 634mL to litres by simply dividing by 1000 because we know 1 litre has 1000ml so 634/1000 = 0.634L

Now we just plug these guys in our molarity formula M=mol/L

M= 0.3035/0.634 = 0.479 M or mol/L

3 0

2 years ago

Read 2 more answers

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago

What is the molar concentration of 35 mL of H2SO4 that neutralizes 25 mL of 0.320M NaOH

stira [4]

V ( H2SO4) = 35 mL / 1000 => 0.035 L

M ( H2SO4) = ?

V ( NaOH ) = 25 mL / 1000 => 0.025 L

M ( NaOH ) = 0.320 M

number of moles NaOH:

n = M x V

n = 0.025 x 0.320 => 0.008 moles of NaOH

Mole ratio:

<span>2 NaOH + H2SO4 = Na2SO4 + 2 H2O
</span>
2 moles NaOH ---------------------- 1 mole H2SO4
0.008 moles moles NaOH ---------- ??

0.008 x 1 / 2 => 0.004 moles of H2SO4 :

Therefore:

M ( H2SO4) = n / V

M = 0.004 / 0.035

= 0.114 M

hope this helps!

6 0

3 years ago