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2 years ago

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If you weighed out a small sample of sodium acetate, then dumped it into a beaker of water and stirred it around, which of these

statements would NOT be true? a. The the sodium acetate will fail to dissolve b. The solution will conduct electricity. c. Individual sodium and acetate ions are present. d. none of the above

1 answer:

aleksandrvk [35]2 years ago

3 0

Answer:

The statement that isn't true is the a)

Explanation:

The sodium acetate when mixed with water dissociates:

$Na(CH_3COO) \longrightarrow Na^+ + CH_3COO^-$

So we can affirm that this ions will dissolve in water (a polar solvent) and individual ions of sodium and acetate will be present in the solution. Also, the presence of i<u>ons in water increases its electrical conductivity</u>.

Being all that said, the statement that isn't true is the a)

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Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the

Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by $\eta$ .

The given data is as follows.

$\eta$ = 0.82, $T_{1}$ = (21 + 273) K = 294 K

$P_{1}$ = 200 kPa, $P_{2}$ = 1000 kPa

Therefore, calculate the final temperature as follows.

$\eta = \frac{T_{2} - T_{1}}{T_{2}}$

0.82 = $\frac{T_{2} - 294 K}{T_{2}}$

$T_{2}$ = 1633 K

Final temperature in degree celsius = $(1633 - 273)^{o}C$

= $1360^{o}C$

Now, we will calculate the entropy as follows.

$\Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}$

For 1 mole, $\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

It is known that for $NH_{3}$ the value of $C_{v}$ = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

$\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

= $0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}$

= 0.0346 kJ/mol

or, = 34.6 J/mol (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0

2 years ago

20 point, pls help. A 5.50 mole sample of a gas has a volume of 2.50 L. What would the volume be if the amount increased to 11.0

alexandr1967 [171]

Answer:

have to go through it is a great place for you to make me some of the members of your family and I will be happy to help you with this process and if you have any questions or concerns about the dad lowkey or if you need help with the project or if you need help with anything else or just let me know what you need me to do it

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1 year ago

A solution is prepared by adding 6.24 g of benzene (C 6H 6, 78.11 g/mol) to 80.74 g of cyclohexane (C 6H 12, 84.16 g/mol). Calcu

tekilochka [14]

Answer:

$x_B=0.0769$

$m=0.990m$

Explanation:

Hello,

In this case, we can compute the mole fraction of benzene by using the following formula:

$x_B=\frac{n_B}{n_B+n_C}$

Whereas n accounts for the moles of each substance, thus, we compute them by using molar mass of benzene and cyclohexane:

$n_B=6.24g*\frac{1mol}{78.11g}=0.0799mol\\ \\n_C=80.74g*\frac{1mol}{84.16g} =0.959mol$

Thus, we compute the mole fraction:

$x_B=\frac{0.0799mol}{0.0799mol+0.959mol}\\ \\x_B=0.0769$

Next, for the molality, we define it as:

$m=\frac{n_B}{m_C}$

Whereas we also use the moles of benzene but rather than the moles of cyclohexane, its mass in kilograms (0.08074 kg), thus, we obtain:

$m=\frac{0.0799mol}{0.08074kg}=0.990mol/kg$

Or just 0.990 m in molal units (mol/kg).

Best regards.

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2 years ago

Aluminum oxide has a composition of 52.9% aluminum and 47.1% oxygen by mass. if 16.4 g of aluminum reacts with oxygen to form al

Dafna1 [17]

The balanced chemical reaction is written as:

4Al + 3O2 = 2Al2O3

To determine the mass of oxygen gas that would react with the given amount of aluminum metal, we use the initial amount and relate this amount to the ratio of the substances from the chemical reaction. We do as follows:

moles Al = 16.4 g ( 1 mol / 26.98 g ) = 0.61 mol Al
moles O2 = 0.61 mol Al ( 3 mol O2 / 4 mol Al ) = 0.46 mol O2
mass O2 = 0.46 mol O2 ( 32.0 g / mol ) = 14.59 g O2

Therefore, to completely react 16.4 grams of aluminum metal we need a minimum of 14.59 grams of oxygen gas.

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2 years ago

If the actual yield of the reaction was 75% instead of 100%, how many molecules of no would be present after the reaction was ov

artcher [175]

To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:

N₂ + O₂ → 2 NO

Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.

1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present

Number of molecules = 0.07154 mol NO(6.022×10²³ molecules/mol)
<em>Number of molecules = 4.3×10²² molecules NO present</em>

8 0

2 years ago

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