All categories

3 years ago

The number of significant figures in 0.000820 g is?

1 answer:

7 0

There are 3 sig figs. 8 and 2 are obvious. The ending 0 is significant because it is at the end of a decimal number, and doesn't have to be measured.

You might be interested in

A solution contains 42.0 g of heptane (C7H16) and 50.5 g of octane (C8H18) at 25 ∘C. The vapor pressures of pure heptane and pur

Kruka [31]

Answer:

(a) 22.3 torr; 5.6 torr; (b) 27.9 torr; (c) 77.7 % heptane; 23.3 % octane

(d) Heptane is more volatile than octane

Explanation:

We can use Raoult's Law to solve this problem.

It states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction. In symbols,

$p_{i} = \chi_{i} p_{i}^{\circ}$

(a) Vapour pressure of each component

Let heptane be Component 1 and octane be Component 2.

(i) Moles of each component

$n_{1} = \text{42.0 g} \times \dfrac{\text{1 mol}}{\text{100.20 g}} = \text{0.4192 mol}\\n_{2} = \text{50.5 g} \times \dfrac{\text{1 mol}}{\text{114.23 g}} = \text{0.4421 mol}$

(ii) Total moles

$n_{\text{tot}} = 0.4192 + 0.4421 = \text{0.8613 mol}$

(iiii) Mole fractions of each component

$p_{1} = 0.4867 \times 45.8 = \textbf{22.3 torr}\\p_{2} = 0.5133 \times 10.9 = \ \textbf{5.6 torr}$

(iv) Partial vapour pressures of each component

$p_{1} = 0.4867 \times 45.8 = \textbf{22.3 torr}\\p_{2} = 0.5133 \times 10.9 = \textbf{5.6 torr}$

(b) Total pressure

$p_{\text{tot}} = p_{1} + p_{2} = 22.3 + 5.6 = \text{27.9 torr}$

(c) Mass percent of each component in vapour

$\chi_{1} = \dfrac{p_{1}}{p_{\text{Tot}}} = \dfrac{22.3}{27.9} =0.799\\\chi_{2} = \dfrac{p_{2}}{p_{\text{Tot}} }= \dfrac{5.6}{27. 9} =0.201$

The ratio of the mole fractions is the same as the ratio of the moles.

$\dfrac{n_{1}}{n_{2}} = \dfrac{0.799}{0.201}$

If we have 1 mol of vapour, we have 0.799 mol of heptane and 0.201 mol of octane

$m_{1} = 0.799 \times 100.20 = \text{80.1 g}\\m_{2} = 0.201\times 114.23 = \text{23.0 g}\\m_{\text{tot}} = 80.1 + 23.0 = \text{103.1 g}\\\\\text{ mass percent heptane} = \dfrac{80.1}{103.1} \times 100 \, \% = \mathbf{77.7\, \%}\\\\\text{ mass percent octane} = \dfrac{23.0}{103.1} \times 100 \, \% = \mathbf{22.3\, \%}}$

(d) Enrichment of vapour

The vapour is enriched in heptane because heptane is more volatile than octane.

5 0

3 years ago

A student increases the temperature of a 417cm³ balloon from 278k to 231k. Assuming constant pressure, what should the new volum

viva [34]

Charles law states that volume of gas is directly proportional to temperature at constant pressure
V/T = k
where V - volume , T - temperature and k - constant
$\frac{V1}{T1} = \frac{V2}{T2}$
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation

in the question it states that the temperature has been increased from 278 K to 231 K but it should actually be temperature is decreased from 278 K to 308 K
substituting the values in the equation
$\frac{417cm^{3} }{278K} = \frac{V}{308 K}$
V = 462 cm³
the answer should be D. 462 cm³

3 0

2 years ago

Identify the intermolecular attractions for dimethyl ether and for ethyl alcohol. Which molecule is expected to be more soluble

zheka24 [161]

Answer:

See explanation

Explanation:

All molecules possess the London dispersion forces. However London dispersion forces is the only kind of intermolecular interaction that exists in nonpolar substances.

So, the only kind of intermolecular interaction that exists in dimethyl ether is London dispersion forces.

As for ethyl alcohol, the molecule is polar due to the presence of polar O-H bond. In addition to London dispersion forces, dipole-dipole interactions and specifically hydrogen bonding also occurs between the molecules.

Because ethyl alcohol is polar, it is more soluble in water than dimethyl ether.

3 0

3 years ago

How many mole ratios can be correctly obtained from the chemical equation 2Al2O3(l) ® 4Al(s) + 3O2(g)?

scoundrel [369]

Answer:

Explanation:

3 0

2 years ago

Aluminium + copper sulfate : Why is this reaction called a displacement reaction

FinnZ [79.3K]

Answer:

Explanation:

6 0

2 years ago