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fredd [130]
2 years ago
5

Determine the volume of a 75-gram sample of gold at stp

Chemistry
1 answer:
Anuta_ua [19.1K]2 years ago
7 0
Let V = the volume of the sample at STP.

The mass is  m = 75 g
The density of gold is 19.32 g/cm³

Because density = mass/volume, therefore
(19.32 g/cm³) = (75 g)/(V cm³)
V = (75 g)/(19.32 g/cm³) = 3.882 cm³

Answer: 3.882 cm³
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consideras util conocer las propiedades extensivas e intensivas de los insumos utilizados para la elaboración de producto ¿por q
Brums [2.3K]

Answer:

Explanation:

No.

Las propiedades físicas de los materiales y sistemas a menudo se pueden clasificar como intensivas o extensivas, según cómo cambia la propiedad cuando cambia el tamaño (o extensión) del sistema. Según la IUPAC, una cantidad intensiva es aquella cuya magnitud es independiente del tamaño del sistema, mientras que una cantidad extensiva es aquella cuya magnitud es aditiva para los subsistemas. Esto refleja las ideas matemáticas correspondientes de media y medida, respectivamente.

Una propiedad intensiva es una propiedad a granel, lo que significa que es una propiedad física local de un sistema que no depende del tamaño del sistema o de la cantidad de material en el sistema. Los ejemplos de propiedades intensivas incluyen temperatura, T; índice de refracción, n; densidad, ρ; y dureza de un objeto.

Por el contrario, propiedades extensivas como la masa, el volumen y la entropía de los sistemas son aditivas para los subsistemas porque aumentan y disminuyen a medida que crecen y se reducen, respectivamente.  

Estas dos categorías no son exhaustivas, ya que algunas propiedades, físicas no son exclusivamente intensivas ni extensivas. Por ejemplo, la impedancia eléctrica de dos subsistemas es aditiva cuando, y solo cuando, se combinan en serie; mientras que si se combinan en paralelo, la impedancia resultante es menor que la de cualquiera de los subsistemas.

¡Espero haberte ayudado!  :)

7 0
2 years ago
5. Which of the following atoms has the largest atomic
Airida [17]

Answer:

b. chlorine

Explanation:

it has the highest atomic radius

4 0
2 years ago
List the number of each type of atom on the left side of the equation C5H12(g)+8O2(g)→5CO2(g)+6H2O(g) C 5 H 12 ( g ) + 8 O 2 ( g
Gala2k [10]

Answer:

Carbon=5, hydrogen=12, oxygen=16

Explanation:

Carbon=5, hydrogen=12, oxygen=16

In order to effectively count the number of atoms, we look at the equation closely and take note of the stoichiometric coefficients of each reactant as this influences the number of atoms of that element present.

For instance, oxygen is diatomic and has a stoichiometric coefficient of 8. This implies the there are sixteen atoms of oxygen altogether.

Note that the left hand side refers to the reactants side.

5 0
2 years ago
Read 2 more answers
The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in diethyl et
Molodets [167]

Answer:

Explanation:

The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in diethyl ether is aspirin. Calculate the vapor pressure of the solution at 25 °C when 14.88 grams of aspirin, C9H8O4 (180.1 g/mol), are dissolved in 269.2 grams of diethyl ether. diethyl ether = CH3CH2OCH2CH3 = 74.12 g/mol.

mol of C4H10O = mass of C4H10O / molar mass of C4H10O

= 242.1 g / 74.12 g/mol

= 3.266 mol

mol of C9H8O4 = mass of C9H8O4 / molar mass of C9H8O4

= 10.33 g / 180.1 g/mol

= 0.05736 mol

mole fraction of C4H10O,

X = mole of CHH1O0 / total mol

= (3.266)/(3.266 + 0.05736)

= 0.9827

now use:

P = Po*X

P = 463.57 * 0.9827

= 455.6 mm Hg

3 0
2 years ago
A sample of 0.6760 g of an unknown compound containing barium ions (ba2+) is dissolved in water and treated with an excess of na
notka56 [123]

Answer: 35.72 % of Barium ions will be present in the original unknown compound.

Explanation: The reaction of Barium ions and sodium sulfate is:

Na_2SO_4(aq.)+Ba^{2+}(aq.)\rightarrow BaSO_4(s)+2Na^+(aq.)

Here, Sodium sulfate is present in excess, Barium ions are the limiting reagent because it limits the formation of product.

Now, 1 mole of barium sulfate is produced by 1 mole of Barium ions.

Molar mass of Barium sulfate = 233.38 g/mol

Molar mass of Barium ions = 137.327 g/mol

233.38 g/mol of barium sulfate will be produced by 137.323 g/mol of Barium ions, so

0.4105 grams of barium sulfate will be produced by = \frac{137.327g/mol}{233.38g/mol}\times 0.4105g of Barium ions

Mass of barium ions = 0.2415 grams

To calculate percentage by mass, we use the formula:

\% mass=\frac{\text{Mass of solute (in grams)}}{\text{Total mass of the solution(in grams)}}\times 100

Mass of the solution = 0.6760 grams

Putting the value in above equation, we get

\% \text{ mass of }Ba^{2+}\text{ ions}=\frac{0.2415g}{0.6760g}\times 100

% mass of Barium ions = 35.72%.

8 0
2 years ago
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