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2 years ago

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A 0.100 m solution of which one of the following solutes will have the highest vapor pressure? A 0.100 m solution of which one o

f the following solutes will have the highest vapor pressure? KClO4 Ca(ClO4)2 NaCl sucrose Al(ClO4)3

1 answer:

Assoli18 [71]2 years ago

5 0

Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

Vapor pressure is inversely proportional to the number of solute particles. Hence, more will be the solute particles lower will be the vapor pressure and vice-versa.

(a) $KClO_{4} \rightarrow K^{+} + ClO^{-}_{4}$

It dissociates to give two particles.

(b) $Ca(ClO_{4})_{2} \rightarrow Ca^{2+} + 2ClO^{-}_{4}$

Total number of particles it give upon dissociation are 1 + 2 = 3. Hence, it gives 3 particles.

(c) $Al(ClO_{4})_{3} \rightarrow Al^{3+} + 3ClO^{-}_{4}$

Total number of particles it give upon dissociation are 1 + 3 = 4. Hence, it gives 4 particles.

(d) Surcose being a cobvalent compound doe not dissociate into ions. Therefore, there will be only 1 particle is present.

(e) $NaCl \rightarrow Na^{+} + Cl^{-}$

Total number of particles it give upon dissociation are 1 + 1 = 2. Hence, it gives 2 particles.

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The volume of a gas at 7.00°c is 49.0 ml. if the volume increases to 74.0 ml and the pressure is constant, what will the tempera

marshall27 [118]

Using charles law
v1/t1=v2/t2
v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2 cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius

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2 years ago

Which statement accurately describes dark matter?

Mandarinka [93]

Answer:

C. It does not emit electromagnetic radiation.

Explanation:

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2 years ago

Read 2 more answers

Cacodyl, which has an intolerable garlicky odor and is used in the manufacture of cacodylic acid, a cotton herbicide, has a mass

Papessa [141]

Answer:

The molecular formula of cacodyl is C₄H₁₂As₂.

Explanation:

<u>Let's assume we have 1 mol of cacodyl</u>, in that case we'd have 209.96 g of cacodyl and the<u> following masses of its components</u>:

209.96 g * 22.88/100 = 48.04 g C

209.96 g * 5.76/100 = 12.09 g H

209.96 g * 71.36/100 = 149.83 g As

Now we convert those masses into moles:

48.04 g C ÷ 12 g/mol = 4.00 mol C

12.09 g H ÷ 1 g/mol = 12.09 mol H

149.83 g As ÷ 74.92 g/mol = 2.00 mol As

Those amounts of moles represent the amount of each component in 1 mol of cacodyl, thus, the molecular formula of cacodyl is C₄H₁₂As₂.

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2 years ago

Determine the molar mass of freon–11 gas if a sample weighing 0.597 g occupies 100. cm3 at 95°c, and 1,000. mmhg (r = 0.08206 l

Stells [14]

We can first find the number of moles using the ideal gas law equation,
PV = nrT
where P - pressure - 1000 mmHg / 760 mmHg/atm = 1.32 atm
V - volume - 100 x 10⁻³ L
n - number of moles
r - universal gas constant - 0.08206 LatmK⁻¹mol⁻¹
T - temperature in Kelvin - 95 °C + 273 = 368 K
substituting these values
1.32 atm x 100 x 10⁻³ L = n x 0.08206 LatmK⁻¹mol⁻¹ x 368 K
n = 0.00437 mol

molar mass can be determined as follows
molar mass = mass present / number of moles
molar mass = 0.597 g / 0.00437 mol = 136.6 g/mol
molar mass of gas is 137 g/mol

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2 years ago

A. Calculations for the Determination of Ammonium Chloride The data from the data entry portion of the report has been copied in

Vladimir79 [104]

Answer:

A

Explanation:

Considering question A

Mass of original sample is $m_o = 0.945 \ g$

Mass of NH4Cl is $m_n = 0.116 \ g$

Percent of NH4Cl is $k = 12.275 \%$

B

Mass of NaCl $m_k = 0.359 \ g$

C

Mass of SiO2 $m_e = 0.46$

D

Mass of original sample $m_o = 0.945 \ g$

Differences in these weights (g) (use the absolute value of the difference)

recovery of matter $G = 0.01 \ g$

The correct option is C

From the question we are told that

The mass of evaporating dish on #1 is $m_1 = 38.646 \ g$

The mass of evaporating dish and original sample $m_2 = 39 591 \ g$

The mass of evaporating dish after subliming $NH_4Cl$ is $m_3 = 39.4750 \ g$

Generally the mass of the original sample is mathematically represented as

$m_o = m_2 - m_1$

=> $m_o = 39 591 - 38.646$

=> $m_o = 0.945 \ g$

Generally the mass of $NH_4Cl$ is mathematically represented as

$m_n = m_2 - m_3$

=> $m_n = 39 591 - 39.4750$

=> $m_n = 0.116 \ g$

The Percent $NaH_4 Cl (g)$

$k = \frac{ m_n}{m_o} *100$

=> $k = \frac{0.116 }{0.945} *100$

=> $k = 12.275 \%$

Considering question B

The mass of evaporating dish #2 is $m_g = 38700\ g$

The mass of watch glass is $m_a = 28 299 \ g$

The mass of evaporating dish #2, watch glass and NaCl $m_b = 67,355 \ g$

Generally the mass of NaCl is

$m_k = m_b -[m_g + m_a]$

=> $m_k = 67,355 -[38700 + 28 299]$

=> $m_k = 0.359 \ g$

Considering question C

The mass of evaporating dish is $m_p= 38.645$

The mass of evaporating dish and SiO2 $m_s = 39.105 \ g$

Generally the mass of SiO2 is mathematically represented as

$m_e = 39.105 - 38.645$

=> $m_e = 0.46$

Considering D

The mass of the original sample is $m_o = 0.945 \ g$

Generally the experimental mass recovered (NH_4Cl,NaCl, SiO2 ) is mathematically evaluated as

$M =0.116 + 0.46 + 0.359$

$M = 0.935 \ g$

Generally the differences in these weights (g) of recovery of matter is mathematically represented as.

$G =0.945- 0.935$

=> $G = 0.01 \ g$

While drying the NaCl, the liquid boiled and some splattered out of the evaporating dish, causing the recovered mass to be lower.

6 0

2 years ago