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2 years ago

What is the mass (in mg) of 2.63 moles of Nickel

Chemistry

2 answers:

Naya [18.7K]2 years ago

7 0

Hope this helps you.

Helga [31]2 years ago

3 0

Times 2.63 by the molar mass (from periodic table) which will give you the answer in grams. Then convert that to mg (10^3).

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For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because _

nata0808 [166]

Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

$\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g)$ .

When $\rm H_2$ is added to $\rm H_2C\text{=}CH_2$ (ethene) under heat and with the presence of a catalyst, $\rm H_3C\text{-}CH3$ (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, $\Delta S < 0$ .

The equation for the change in Gibbs Free Energy for a particular reaction is:

$\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}})$ .

For a particular reaction, the more negative $\Delta G$ is, the more spontaneous ("favorable") the reaction would be.

Since typically $\Delta S < 0$ for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

$T$ (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of

8 0

3 years ago

A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.62 g of water at 52.3 oC in an insulated container. clear = 0.128

alisha [4.7K]

Answer: The final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

Explanation:

The given data is as follows.

mass = 7.62 g, $T_{2} = 10.8^{o}C$

Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.

q = $mC \times \Delta T$

= $7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$

Now, heat gained by lead will be calculated as follows.

q = $mC \times \text{Temperature change of lead}$

= $2.04 \times 0.128 \times (T - 11.0)$

According to the given situation,

Heat lost = Heat gained

$7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$ = $2.04 \times 0.128 \times (T - 11.0)$

T = $50.26^{o}C$

Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

7 0

3 years ago

What is the mass of 0.500 mole of Ba? (Watch sf’s) 68.7 g 137.3 g 68.65 g 69 g

Contact [7]

68.7g

molar mass of Ba(137.327)x (.500)= 68.6635

.500 has 3 sig figs.

8 0

2 years ago

Nvm i dont have any questions

gregori [183]

Answer:

well if you ever do neeed help im here

Explanation:

5 0

2 years ago

If 50 liters of carbon dioxide gas are produced by the reaction below at 273 Kand 100 atm, how oxygen gas were used the reaction

oee [108]

Using the ideal gas equation:

pV = nRT
n = pV / RT

1atm = 101325Pa, so p = 10132500Pa
1L = 0.001m^3, so V = 0.050m^3
R = 8.214 (ideal gas constant)
T = 273K

Hence moles of CO2 = (10132500 * 0.050) / (8.314 * 273) = 223.2101553

Reaction ratio between oxygen and CO2 is 1:2
Hence moles of O2 = 223... / 2 = 112 moles (3sf)

5 0

3 years ago