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3 years ago

5

Four 22.4 samples of nitrogen are bottled at STP and placed into environments with varying temperature. Which environment yield

a sample the the greatest kinetic energy?
A)100k
B) 303k
C)273 k
D)200k

1 answer:

Paha777 [63]3 years ago

7 0

Mayeb a is the correct answer .....

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PLEASE HELP!!!!!!!

dsp73

Answer: 1. $2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu$

2. 3 moles of $CuCl_2$ : 2 moles of $Al$

3. 0.33 moles of $CuCl_2$ : 0.92 moles of $Al$

4. $CuCl_2$ is the limiting reagent and $Al$ is the excess reagent.

5. Theoretical yield of $AlCl_3$ is 29.3 g

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Al=\frac{25.0g}{27g/mol}=0.92moles$

$\text{Moles of} CuCl_2=\frac{45.0g}{134g/mol}=0.33moles$

The balanced chemical equation is:

$2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu$

According to stoichiometry :

3 moles of $CuCl_2$ require = 2 moles of $Al$

Thus 0.33 moles of $CuCl_2$ will require= $\frac{2}{3}\times 0.33=0.22moles$ of $Al$

Thus $CuCl_2$ is the limiting reagent as it limits the formation of product and $Al$ is the excess reagent.

As 3 moles of $CuCl_2$ give = 2 moles of $AlCl_3$

Thus 0.33 moles of $CuCl_2$ give = $\frac{2}{3}\times 0.33=0.22moles$ of $AlCl_3$

Theoretical yield of $AlCl_3=moles\times {\text {Molar mass}}=0.22moles\times 133.34g/mol=29.3$

Thus 29.3 g of aluminium chloride is formed.

5 0

3 years ago

When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for

balu736 [363]

Answer:

1.

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

$Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-$

Whereas the half reactions are:

$Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O$

Next, we exchange the transferred electrons:

$1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O$

Afterwards, we add them to obtain:

$Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O$

By adding and subtracting common terms we obtain:

$Sn^0+4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O$

Finally, by removing the oxidation states we have:

$Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O$

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

6 0

3 years ago

Formula of sodium bicarbonate<br>please help me with this

Stels [109]

Answer:

NaHCO₃

Explanation:

Sodium bicarbonate (baking soda) is a chemical compound with the formula NaHCO₃.

5 0

3 years ago

Which of the following is a component of cell theory that is supported by the diagram

Crazy boy [7]

The one on top is corrrect because the cell diagram will always contain one object or more

8 0

4 years ago

1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.

Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice ) = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

$\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}$

$\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}$

$\Delta \ S = {(1*37.7)In \dfrac{263}{273}$

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0

3 years ago