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tankabanditka [31]

3 years ago

7

Indica el período, el grupo, el nombre y el número atómico de los elementos que se representan con las siguientes configuracione

s electrónicas:

1 answer:

motikmotik3 years ago

4 0

Answer:

have g8igcicg9ccgco

Explanation:

gcohogxzrxog floor. gi gi gi gi. r. uvgi7rz. g yde6ifkg. gg g goog oh g of ifz8fd8dti

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How does sustainable development support in a long term vision

vfiekz [6]

Explanation:

meet .google .com/sxr-wgwg-vnc

4 0

3 years ago

Read 2 more answers

Answer the question briefly: How can atom collapse and why are atoms stable?

kykrilka [37]

The nuclei of atoms become unstable when the repelling forces of the protons cannot be balanced by the number of neutrons in the nucleus. It then re-arranges itself randomly to a more stable configuration by emitting any of a series of particles. During radioactive decay, an atom does not collapse.

Since an atom is mostly empty space - that is it’s nucleus is relatively distant from the electron shells so, in the presence of extreme forces such as gravity inthe collapse of a large star, the inward pressures on the atom overcome the natural balance of the atomic structure and the ‘empty space’ disappears as nuclei are mashed together by the intense pressures and a neutron star is formed. Under even more external pressure, even the neutron star can collapse to form a black hole.

8 0

4 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$

Hence, the freezing point of solution is 2.6°C

8 0

3 years ago

Salivary amylase is an enzyme produced by the salivary glands that breaks down carbohydrates. What will happen to this enzyme as

o-na [289]

Answer:

It will be denatured.

Explanation:

Salivary amylase works best at pH 6.8.

At pH 2.5, its activity will decrease enormously as it becomes denatured by the high acidity of the stomach contents.

4 0

3 years ago

How much energy is required to raise the temperature of 10.6 grams of gaseous neon from

Alona [7]

Answer:

Approximately $1.95 \times 10^{2}\; \rm J$ .

Explanation:

Look up the specific heat of gaseous neon:

$c = 1.03 \; \rm J \cdot g^{-1} \cdot K^{-1}$ .

Calculate the required temperature change:

$\Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K$ .

Let $m$ denote the mass of a sample of specific heat $C$ . Energy required to raise the temperature of this sample by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ .

For the neon gas in this question:

$c = 1.03\; \rm J \cdot g^{-1}\cdot K^{-1}$ .
$m = 10.6\; \rm g$ .
$\Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K$ .

Calculate the energy associated with this temperature change:

$\begin{aligned}Q &= c \cdot m \cdot \Delta T \\ &= 1.03\; \rm J \cdot g^{-1}\cdot K^{-1} \times 10.6\; \rm g \times 17.9\; \rm K \\ &\approx 1.95 \times 10^{2}\; \rm J\end{aligned}$ .

3 0

3 years ago