Now can you answer pls

How many magnesium atoms are contained in 3.75 moles of magnesium?

Katarina [22]

In 1 mole of magnesium there are Avogadro's number of atoms are present.
Avogadro's number = 6.023 x 10²³
1 mole = 6.023 x 10²³ atoms
3.75 moles = 3.75 x 6.023 x 10²³
=2.26 x 10²⁴
So, in 3.75 moles of magnesium there are 2.26 x 10²⁴ atoms of magnesium are present.

7 0

3 years ago

A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad

ASHA 777 [7]

$0 \; \textdegree{\text{C}}$

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

$E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ}$ and
$m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}$

Let the final temperature of the system be $t \; \textdegree{\text{C}}$ . Thus $\Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial}) - t_{0} = t \; \textdegree{\text{C}}$

$Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ}$ (converted to kilojoules)
$Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}$
$Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}$

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable $t$ .

$E(\text{absorbed} ) = E(\text{released})$

$E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})$
$E(\text{released}) = Q(\text{lead})$

Confirm the uniformity of units, equate the two expressions and solve for $t$ :

$66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)$

$t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}}$ which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., $t \le 0 \; \textdegree{\text{C}}$ . However, there's no way for the temperature of the system to go below $0 \; \textdegree{\text{C}}$ ; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at $0 \; \textdegree{\text{C}}$ ; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

3 0

3 years ago

Given the following reaction: NH4SH (s) <--> NH3 (g) + H2S (g) If we start

almond37 [142]

Answer:

D. 0.3 M

Explanation:

NH4SH (s) <--> NH3 (g) + H2S (g)

Initial concentration 0.085mol/0.25L 0 0

Change in concentration -0.2M +0.2 M +0.2M

Equilibrium 0.035mol/0.25 L=0.14M 0.2M 0.2M

concentration

Change in concentration (NH4SH) = (0.085-0.035)mol/0.25L =0.2M

K = [NH3]*[H2S]/[NH4SH] = 0.2M*0.2M/0.14M ≈ 0.29 M ≈ 0.3M

4 0

3 years ago

Can someone help me?

Natalka [10]

Answer 19.9g. I’ve took the test last week at my uncle randy’s house

8 0

3 years ago

This is a type of element that has many valence electrons, not a conductor.

cestrela7 [59]

What is the question

3 0

3 years ago