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miss Akunina
1 month ago
5

Every single-celled organism is able to survive because it carries out *

Chemistry
1 answer:
castortr0y [3K]1 month ago
6 0

Every unicellular organism prospers by executing metabolic activities.

Metabolic activities encompass the set of chemical reactions essential for sustaining life.

Explanation:

Different metabolic pathways maintain an organism's viability. Various metabolic activities occur in all living organisms.

These include processes like cellular respiration, reproduction, excretion, and digestion. Each living cell engages in these activities to survive.

Organisms acquire the energy necessary for these activities through food consumption.

Read more on -

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26.2 g piece of copper metal is heated from 21.5°C to 201.6°C. Calculate the amount of heat absorbed by the metal. The specific
lorasvet [2795]

Response:1816.6 joules

Clarification:refer to the attached image

6 0
21 day ago
A stock solution of Cu2+(aq) was prepared by placing 0.8875 g of solid Cu(NO3)2∙2.5 H2O in a 100.0-mL volumetric flask and dilut
Anarel [2989]

Answer:

3.816 × 10⁻³ M

Explanation:

A stock solution of Cu²⁺(aq) is made by dissolving 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask, and then brought up to volume with water. What is the molarity (in M) of Cu²⁺(aq) in this stock solution?

We can derive the following relations:

  • The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
  • Each mole of Cu(NO₃)₂∙2.5H₂O yields one mole of Cu²⁺.

The moles of Cu²⁺ present in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:

0.8875gCu(NO_{3})_{2}.2.5H_{2}O\times \frac{1molCu(NO_{3})_{2}.2.5H_{2}O}{232.59gCu(NO_{3})_{2}.2.5H_{2}O} \times \frac{1molCu^{2+} }{1molCu(NO_{3})_{2}.2.5H_{2}O} =3.816\times10^{-3} molCu^{2+}

The molarity of Cu²⁺ is:

\frac{3.816\times10^{-3} mol}{100.0 \times10^{-3}L} =3.816\times10^{-2}M

4 0
1 month ago
How many molecules of PF5 are found in 39.5 grams of PF5?
Anarel [2989]

Response:

1.9 \times 10^{23} molecules of PF_5 can be found in 39.5 grams of PF_5.

Clarification:

Atomic weights: P= 31, F= 19,

The molar mass equals 1 atomic weight of P + 5 atomic weights of

 F= 31+5 × 19\times = 31+95

=126 g/mole

The number of moles in 39.5 gm of

equals \frac{Mass}{Molar mass}

 = \frac{39.5}{126}moles

 =0.3134 moles

1 mole of any substance encompasses

0.3131 moles comprises 0.3134

= 1.9 \times 10^{23} molecules

Thus, 1.9 \times 10^{23} molecules of PF_5 can be found in 39.5 grams of PF_5.

7 0
12 days ago
At the boiling point, the density of the liquid is 809 g/l and that of the gas is 4.566 g/l. how many liters of liquid nitrogen
KiRa [2933]

Result: 1.68 L of liquid nitrogen is generated during the gas liquefaction process.

Clarification:

This process involves transforming gaseous nitrogen into its liquid form.

The two states possess distinct densities, thus occupying varying volumes; however, the mass remains constant.

Step 1: Calculate the mass of nitrogen gas

Let’s determine the mass of nitrogen gas associated with 297 L.

The density formula is:

Density = \frac{Mass}{Volume}

With a density of nitrogen gas at 4.566 g/L and a volume of 297 L, we can compute the mass of nitrogen gas as follows:

Using these values yields:

Mass = Density \times Volume

Mass = \frac{4.566g}{L} \times 297L

Mass = 1356g

The mass of nitrogen gas calculates to be 1356 g.

Step 2: Derive the volume of liquid nitrogen from the mass obtained

The mass for liquid nitrogen remains the same.

With the density of liquid nitrogen at 809 g/L, we can substitute this into our formula to find the volume of liquid.

Volume = \frac{Mass}{Density}

Volume = \frac{1356g}{809g/L}

Therefore, the volume of liquid nitrogen is 1.68 L.


6 0
1 month ago
In a titration experiment, H2O2(aq) reacts with aqueous MnO4-(aq) as represented by the equation above. The dark purple KMnO4 so
Alekssandra [3086]

Respuesta:

El oxígeno en H2O2 es la especie que se reduce a H2O y se oxida a O2.

Explicación:

5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g)

La oxidación se define como la pérdida de electrones. La oxidación provoca un aumento en el número de oxidación de un elemento.

Si se descompone esta reacción en sus mitades de reducción y oxidación

Se observa que, de los reactivos mencionados anteriormente,

H202 se convierte en H2O y O2

MnO4- + H+ se convierte en Mn2+ y H2O

El número de oxidación de Mn cambia de +7 en MnO4- a +2 en Mn2+ (lo que indica evidentemente una reducción)

El oxígeno en MnO4- no cambia su número de oxidación, ya que se mantiene en -2

El número de oxidación del oxígeno cambia de -1 en H2O2 a -2 en H2O y 0 en O2

El hidrógeno en H2O2 no cambia su número de oxidación, y su número de oxidación se mantiene en +1 tanto en H2O2 como en H2O.

Esto indica que H2O2 sufre tanto oxidación como reducción; más específicamente, el oxígeno en H2O2 es la especie que se reduce a H2O y se oxida a O2.

Espero que esto ayude

7 0
1 month ago
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