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3 years ago

Which statement is TRUE regarding the macroscopic and

Chemistry

1 answer:

damaskus [11]3 years ago

7 0

Answer:

Chemists make observations on the macroscopic a scale that lead to conclusions about microscopic features

Explanation:

Many important chemical observations are made on the macroscopic scale. This is because, many of the scientific equipments available are not presently able to provide direct evidence about microscopic processes. Evidences obtained from macroscopic observations could serve as important insights into the nature of certain microscopic processes.

This is evident in the study of the structure of the atom. Most of the evidences that led to the deduction of the atomic structure were obtained from macroscopic evidence but ultimately provided important information about the microscopic structure of the atom.

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A reaction takes place that results in an amount of 585 grams of product that is recovered. The theoretical amount calculates to

lisov135 [29]

Answer:

90%

Explanation:

Data obtained from the question include:

Actual yield = 585g

Theoretical yield = 650g

%yield =?

%yield = Actual yield /Theoretical yield x100

% yield = 585/650 x 100

% yield = 90%

7 0

4 years ago

An organic compound contains , , , and . Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2587 g and 0.0861 g .

Kisachek [45]

The question displayed below shows the missing information which therefore completes the question.

An organic compound contains C, H, N and O. Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2587 g of CO2 and 0.0861 g of H2O. A sample of 0.4831 g of the compound was analyzed for nitrogen by the Dumas method. The compound is first reacted by passage over hot: The product gas is then passed through a concentrated solution of to remove the. After passage through the solution, the gas contains and is saturated with water vapor. At STP, 38.9 mL of dry N2 was obtained. In a third experiment, the density of the compound as a gas was found to be 2.86 g/L at 127°C and 256 torr. What are the empirical and molecular formulas of the compound? (Enter the elements in the order: C, H, N, O.)

Answer:

the empirical formula = $\mathbf {C_3H_6O_{12}N}$

the molecular formula = $\mathbf {C_3H_6O_{12}N}$

Explanation:

From the given information:

$\bigg ( 0.2587 \ g \ of CO_2 \bigg) \times \dfrac{1 \ mol \ of CO_2}{44 \ of \ CO_2} \times \dfrac{1 \ mol \ of \ C}{1 \ mol \ of CO_2}$

$= 0.00588 \ mol \ of \ C \times \dfrac{12.01 \ g \ of \ C}{1 \ mol \ of \ C }$

= 0.0706g of C

$\bigg ( 0.0861\ g \ of H_2O \bigg) \times \dfrac{1 \ mol \ of H_2O}{18.02 \ g \ of \ H_2O} \times \dfrac{2 \ mol \ of \ H}{1 \ mol \ of H_2O}$

$=0.0096 \ mol \times \dfrac{1.008 \ g \ of \ H}{1 mol \ H}$

0.0097g of H

Given that N2 at STP = 1 atm, 273 K and V = 0.0389 L

PV = nRT

n = PV/RT

$n = \dfrac{1 \ atm \times 0.0389 \ of \ H_2}{0.0821 \ L.atm /mol.K \times 273 \ K }$

n = 0.00173 mol of N2

The oxygen in the sample = The total grams in sample - gram in H - gram in C

The oxygen in the sample = 0.1023 g - 0.0097 g - 0.706 g

The oxygen in the sample = 0.022 g of O

The number of moles of $O_2 = \dfrac{0.02}{16}$

= 0.001375 mol of O

$O \ in \ product = (0.00588 \ mol \ of \ C ) \times \dfrac{2 \ mol \ of \ O }{1 \ mol \ of \ C }+ \bigg ( 0.0096 \ mol \ of \ H ) \times \dfrac{1 \ mol \ of \ O }{1 \ mol \ of \ H}$

O in product = 0.02136 mol of O

∴

we are meant to divide the moles of each compound by the smallest number of moles; we have:

$C = \dfrac{0.00588}{0.00173} \simeq 3$

$H = 0.0096 = \dfrac{0.0096}{0.00173} \simeq 6$

$O = 0.0199= \dfrac{0.0199}{0.00173} \simeq 12$

$N = 0.00173= \dfrac{0.00173}{0.00173} \simeq 1$

Thus; the empirical formula = $\mathbf {C_3H_6O_{12}N}$

To estimate the molecular formula; we have:

$MM = \dfrac{dRT}{P}$

$MM = \dfrac{2.80 \ g/ L \times 0.0821 \ L.atm /mol.K \times 400 \ K }{0.337 \ atm}$

MM = 272.86 g/mol

Also; the molar mass of $\mathbf {C_3H_6O_{12}N}$ = 248 g/mol

∴

$= \dfrac{272.86 \ g/mol}{248 \ g/mol}$

$=1$

Thus; we can conclude that empirical formula as well as the molecular formula are the same.

6 0

3 years ago

Which best describes a scientist who solves a new problem using something already learned?

adelina 88 [10]

Answer:

The answer should be Skeptical

6 0

3 years ago

In one scene in the movie The Godfather II,

Ad libitum [116K]

10 cm in a cube to find volume is 10 by 10 by 10
1000 cm cubed

10 cm is 0.1 m
0.1 times 0.1 times 0.1 is 0.001 metres cubed
19,300 times 0.001 is
19.3 kg
approximately 193newtons
yes it could be passed around casually

5 0

3 years ago

If 4.65 LL of CO2CO2 gas at 22 ∘C∘C at 793 mmHg mmHg is used, what is the final volume, in liters, of the gas at 35 ∘C∘C and a p

otez555 [7]

Answer:

About 7.9 L.

Explanation:

We can utilize the ideal gas law. Recall that:

$\displaystyle PV = nRT$

Because the amount of carbon dioxide does not change, we can rearrange to formula to:
$\displaystyle \frac{PV}{T}= nR$

Because the right-hand side stays constant, we have that:
$\displaystyle \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = nR$

Hence substitute initial values and known final values:
$\displaystyle \begin{aligned} \frac{(793\text{ mm Hg})(4.65 \text{ L})}{(22 \text{ $^\circ$C})} & = \frac{(743 \text{ mm Hg})V_2}{(35\text{ $^\circ$C})} \\ \\ V_2 & = 7.9\text{ L}\end{aligned}$

Therefore, the final volume is about 7.9 L.

8 0

2 years ago