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3 years ago

Which of these solutes raises the boiling point of water the most?

1 answer:

4 0

Raising of the boiling point is a colligative property. That means that it depends on the number of particles dissolved. The greater the number of particles the greater the increase in the boiling point. So, you can compare the effect of these solutes in the increase of the boiling point by writing the chemical equations and comparing the number of particles dissolved: 1)ionic lithium chloride, LiCl(s) --> Li(+) + Cl (-) => 2 ions; 2) ionic sodium chloride, NaCl(s) --> Na(+) + Cl(-) => 2 ions; 3) molecular sucrose, C12H22O11 (s) ---> C12H22O11(aq) => 1 molecule; 4) ionic phosphate, Na3PO4 --> 3Na(+) + PO4 (3-) => 4 ions; 5) ionic magnesium bromide, MgBr2 --> Mg(2+) + 2 Br(-) => 3 ions. So, ionic phosphate produces the greatest number of particles and it will cause the greatest increase of the boiling point.

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How many grams are in 3.21 x 1024 molecules of potassium hydroxide?

MakcuM [25]

Answer:

3287.04 grams

Explanation:

3 0

3 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

What is the half-life of an isotope that decays to 6.25% of its original activity in 18.9 hours?

inessss [21]

Radioactive material obeys 1st order decay kinetics,
For 1st order reaction, we have
k = $\frac{2.303}{t}Xlog \frac{\text{initial conc.}}{\text{final conc.}}$
where, k = rate constant of reaction

Given: Initial conc. 100, Final conc. = 6.25, t = 18.9 hours

∴ k = $\frac{2.303}{18.9} X log \frac{100}{6.25}$ = 0.1467 hours^(-1)

Now, for 1st order reactions: half life = $\frac{0.693}{k} = \frac{0.693}{0.1467}$ = 4.723 hours.

8 0

3 years ago

When one metal mixes into the crystalline lattice of another it is called a(n) alloy compound chemical reaction heterogeneous so

bekas [8.4K]

the correct answer is alloy

6 0

3 years ago

Hello! Can you please help me with this because I don’t get it at all. Thanks.

denpristay [2]

Answer:

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Explanation:

4 0

3 years ago

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