Select the correct answer.

A sample of nitrogen containing 3.0×10^23 molecules has the same number of molecules as a sample containing

Daniel [21]

Answer:

D) 0.50 mole of Ne

Explanation:

Given data:

Number of molecules of nitrogen = 3.0×10²³ molecules

Which sample contain same number of molecules as nitrogen= ?

Solution:

A) 0.25 mole of O₂

1 mole = 6.022×10²³ molecules

0.25 mol × 6.022×10²³ molecules / 1 mol

1.51×10²³ molecules

B) 2.0 moles of He.

1 mole = 6.022×10²³ molecules

2.0 mol × 6.022×10²³ molecules / 1 mol

12.044×10²³ molecules

C) 1.0 moles of H₂

1 mole = 6.022×10²³ molecules

D) 0.50 mole of Ne

1 mole = 6.022×10²³ molecules

0.50 mol × 6.022×10²³ molecules / 1 mol

3.0×10²³ molecules

5 0

3 years ago

To prevent tank rupture during deep-space travel, an engineering team is studying the effect of temperature on gases confined to

uranmaximum [27]

Answer: 63.88 atm

Explanation:

To answer this, we use the formula PV = nRT since the asumption is that the gas has an ideal behavior

where number of mole = 2.60 mol, R(gas constant) = 0.08205746 L atm/K mol,

T = 251 ∘C = (251 + 273) K = 524 K, Volume = 1.75 L

Making Pressure the subject of the formula, we have

P = nRT/V = 2.6 * 0.08205746 * 524/2.75 = 63.88 atm

6 0

3 years ago

When 1.550 g of liquid hexane (C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87∘C to 38.13∘C.

AlladinOne [14]

Explanation:

1). The given data is as follows.

$T_{i} = 25.87^{o}C$ , $T_{f} = 38.13^{o}C$

C = 5.73 $kJ/^{o}C$

Hence, calculate the change in enthalpy of the reaction as follows.

$\Delta E_{rxn} = -C \times \Delta T$

= $-5.73 \times (38.13 - 25.87)^{o}C$

= -70.25 KJ

As, number of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{1.55}{(6 \times 12 + 14 \times 1)}$

= 0.018 mol

Therefore, enthalpy of reaction in kJ/mol hexane is as follows.

$\Delta E_{rxn} = \frac{-70.25 KJ}{0.018 mol}$

= $-3.90 \times 10^{3}$ kJ/mol

Thus, we can conclude that $\Delta E_{rxn}$ for the reaction in kJ/mol hexane is $-3.90 \times 10^{3}$ kJ/mol .

2). As we know that,

Number of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{1.55}{(7 \times 12 + 8 \times 1)}$

= 0.017 mol

$\Delta E_{rxn} = \E_{rxn} per mol \times \text{number of moles}$

= $-3.91 \times 10^{3} \times 0.017$

= -65.875 kJ

As, $\Delta E_{rxn} = C \times \Delta T
$

-65.875 kJ = $-C \times (37.57 - 23.12)^{o}C$

C = 4.56 $kJ/^{o}C$

Hence, heat capacity of the bomb calorimeter is 4.56 $kJ/^{o}C$ .

3 0

4 years ago

Based on the reactivities of the elements involved, which reaction will form products that are more stable than the reactants?

LUCKY_DIMON [66]

Answer: The correct option is B.

Explanation: All the given reactions are a type of single displacement reactions.

Single displacement reactions are defined as the reactions in which a more reactive metal displaces a less reactive metal in a chemical reaction.

General equation for these reactions is:

$AB+C\rightarrow CB+A$

For the given reactions, we will use the reactivity series. The metal which lies above in the reactivity series will easily displace the metal which lies below in the reactivity series.

Option A: For this reaction, zinc lies below aluminium in the reactivity series. Hence, zinc will not be able to displace aluminium in the chemical reaction.

Option B: For this reaction, Sodium is more reactive than calcium as it lies above in the reactivity series. Hence, sodium will easily displace calcium in the given chemical reaction.

Reaction follows:

$CaBr_2+2Na\rightarrow 2NaBr+Ca$

Option C: For this reaction, hydrogen lies below magnesium in the reactivity series. Hence, hydrogen will not be able to displace magnesium in the chemical reaction.

Option D: For this reaction, Calcium lies below barium in the reactivity series. Hence, calcium will not be able to displace barium in the chemical reaction.

Option E: For this reaction, barium lies below lithium in the reactivity series. Hence, barium will not be able to displace lithium in the chemical reaction.

Reactivity series is attached below.

3 0

4 years ago

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Claire decided to find out which wildflowers are most abundant near roadsides in her community. Which of the following would BES

ruslelena [56]

Answer:

observation

Explanation:

3 0

3 years ago

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