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2 years ago

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Calculate the freezing temperature of the following solution of 0.50 M glucose (a covalent compound). Assume that the molality o

f the solution is 0.50 m. (The molar and molal concentrations of dilute aqueous solutions are often identical to two significant figures.) Enter your answer in the provided box. 0.50 m glucose (a covalent compound) °C

1 answer:

kirza4 [7]2 years ago

4 0

Answer:

-0.93 °C

Explanation:

Hello,

The freezing-point depression is given by:

$T_f-T_f^*=-iK_{solvent}m_{solute}$

Whereas $T_f$ is the freezing temperature of the solution, $T_f^*$ is the freezing temperature of the pure solvent (0 °C since it is water), $i$ the Van't Hoff factor (1 since the solute is covalent), $K_{f,solvent}$ the solvent's freezing point depression point constant (in this case $1.86 C\frac{kg}{mol}$ ) and $m_{solute}$ the molality of the glucose.

As long as the unknown is $T_f$ , solving for it:

$T_f=T_f^*-iK_fm\\T_f=0C-1*1.86C\frac{kg}{mol}*0.5\frac{mol}{kg} \\T_f=-0.93C$

Best regards.

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If 16.9 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc

hodyreva [135]

Answer: 69.72 kg of cryolite will be produced.

Explanation:

The balanced chemical equation is:

$Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of $Al_2O_3$ = $\frac{16.9\times 1000g}{102g/mol}=165.7moles$

moles of $NaOH$ = $\frac{57.4\times 1000g}{40g/mol}=1435moles$

moles of $HF$ = $\frac{57.4\times 1000g}{20g/mol}=2870moles$

As 1 mole of $Al_2O_3$ reacts with 6 moles of $NaOH$

166 moles of $Al_2O_3$ reacts with = $\frac{6}{1}\times 166=996$ moles of $NaOH$

As 1 mole of $Al_2O_3$ reacts with 12 moles of $HF$

166 moles of $Al_2O_3$ reacts with = $\frac{12}{1}\times 166=1992$ moles of $HF$

Thus $Al_2O_3$ is the limiting reagent.

As 1 mole of $Al_2O_3$ produces = 2 moles of cryolite

166 moles of $Al_2O_3$ reacts with = $\frac{2}{1}\times 166=332$ moles of cryolite

Mass of cryolite $(Na_3AlF_6)$ = $moles\times {\text {molar mass}}=332mol\times 210g/mol=69720g=69.72kg$

Thus 69.72 kg of cryolite will be produced.

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2 years ago

How does fertilizer benefit the daily lives of people?

vitfil [10]

Answer:

Fertilizers play an important role in providing crops with the nutrients they need to grow and be harvested for nutritious food. Fertilizers help deliver enough food to feed the world's population. ...

For example, micronutrients such as zinc are important to human nutrition, especially children.

Explanation:

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3 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

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3 years ago

Water (H2O) is made up of molecules with a bent structure, whereas hydrogen sulfide (H2S) is also made up of molecules with a be

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3 years ago

Identify the gas law that applies to the following scenario: If a gas in a closed container is pressurized from 18.0 atm to 14.0

nalin [4]

Answer:

Gay-Lussac's Law

Explanation:

The pressure is directly proportional to the absolute temperature under constant volume. This states the Gay-Lussac's law. The equation is:

P1T2 = P2T1

<em>Where P is pressure and T absolute temperature of 1, initial state and 2, final state of the gas.</em>

<em />

That means the right option is:

- Gay-Lussac's Law

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2 years ago