Why is it important to know which category of microorganism has caused an infection?

Teniendo en cuenta los siguientes fenómenos: ebullición del agua- movimiento de un cuerpo- disolución de sal en agua- combustión

miv72 [106K]

Answer:

Las siguientes son reacciones químicas;

combustión de leña

oxidación del hierro

descomposición del agua en hidrógeno y oxígeno

Explanation:

Una reacción química da como resultado la formación de una (s) sustancia (s) nueva (s), mientras que un cambio físico no conduce a la formación de una sustancia nueva.

Las siguientes son reacciones químicas;

combustión de leña: la combustión de madera implica la oxidación del carbono según la reacción; C (s) + O2 (g) -------> CO2 (g)

oxidación del hierro: La oxidación del hierro conduce a la formación de óxidos de hierro. Como; 2Fe (s) + O2 (g) ----> 2FeO (s)

descomposición del agua en hidrógeno y oxígeno: esta es una reacción química en la que el agua se descompone de la siguiente manera; 2H2O (l) -----> 2H2 (g) + O2 (g)

Todos estos procesos enumerados anteriormente conducen a la formación de nuevas sustancias, por lo tanto, son reacciones químicas.

3 0

2 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

3 years ago

How many molecules are in 100 g of C6H120,?*

Grace [21]

Answer:

3.37 × 10²³ molecules

Explanation:

Given data:

Mass of C₆H₁₂O₆ = 100 g

Number of molecules = ?

Solution:

Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

Number of moles = 100 g/ 180.16 g/mol

Number of moles = 0.56 mol

Number of molecules:

1 mole contain 6.022 × 10²³ molecules

0.56 mol × 6.022 × 10²³ molecules /1 mol

3.37 × 10²³ molecules

6 0

3 years ago

Which of these is the final result of secondadry succession?

Grace [21]

Answer:

a

Explanation:

ADAPTATIOnn but if thhere would be an option o all the above it would be that

5 0

3 years ago

Read 2 more answers

what conclusions can be made about the relationship between metallic character and the atomic radius?

kolezko [41]

We have to get the relationship between metallic character and atomic radius.

Metallic character increases with increase in atomic radius and decrease with decrease of atomic radius.

If electrons from outermost shell of an element can be removed easily, that atom can be considered to have more metallic character.

With increase in atomic radius, nuclear force of attraction towards outermost shell electron decreases which facilitates the release of electron.

With decrease in atomic radius, nuclear force of attraction towards outermost shell electrons increases, so electrons are hold tightly to nucleus. Hence, removal of electron from outermost shell becomes difficult making the atom less metallic in nature.

5 0

3 years ago