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sleet_krkn [62]

3 years ago

9

Stan is a member of the Association for Consumer Research, but when he attends the conferences, he doesn't identify with the gro

up. Even though his research is received favorably, he often feels detached and perceives himself as different from the other members. Stan can best be described as having a(n) _____ self-schema with respect to this group.idealizedseparatedactualizedconnectedambiguous

1 answer:

dolphi86 [110]3 years ago

7 0

Answer:

Separated

Explanation:

Stan does not feel he identifies closely with the group. He feels the other members are close knit with each other but he is not part of the group.

For Stan, it would be an illusion to say he has a close relationship with the members of the group

This infers that Stan has a separated self-schema with respect to the group.

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Automobile airbags contain solid sodium azide, NaN3, that reacts to produce nitrogen gas when heated, thus inflating the bag. 2N

Vitek1552 [10]

Answer : The value of work done for the system is 1144.69 J

Explanation :

First we have to calculate the moles of $NaN_3$

$\text{Moles of }NaN_3=\frac{\text{Mass of }NaN_3}{\text{Molar mass of }NaN_3}$

Molar mass of $NaN_3$ = 65.01 g/mole

$\text{Moles of }NaN_3=\frac{20.2g}{65.01g/mole}=0.311mole$

Now we have to calculate the moles of nitrogen gas.

The balanced chemical reaction is,

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced reaction we conclude that

As, 2 mole of $NaN_3$ react to give 3 mole of $N_2$

So, 0.311 moles of $NaN_3$ react to give $\frac{0.311}{2}\times 3=0.466$ moles of $N_2$

Now we have to calculate the volume of nitrogen gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $N_2$ gas = 1.00 atm

V = Volume of $N_2$ gas = ?

n = number of moles $N_2$ = 0.466 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $22^oC=273+22=295K$

Putting values in above equation, we get:

$1.00atm\times V=0.466mole\times (0.0821L.atm/mol.K)\times 295K$

$V=11.3L$

As initially no nitrogen was present. So,

Volume expanded = Volume of nitrogen evolved

Thus,

Expansion work = Pressure × Volume

Expansion work = 1.00 atm × 11.3 L

Expansion work = 11.3 L.atm

Conversion used : (1 L.atm = 101.3 J)

Expansion work = 11.3 × 101.3 = 1144.69 J

Therefore, the value of work done for the system is 1144.69 J

5 0

3 years ago

stearic acid and linoleic acid stearic acid and linoleic acid stearic acid has the higher melting point, because it has two more

elena55 [62]

Acid palmitic acid has higher melting point, because it has two more methylene groups.

$\text { Myristic acid } \rightarrow \mathrm{CH}_3\left(\mathrm{CH}_2\right)_{12} \mathrm{COOH} \\& \text { Palmitic acid } \rightarrow \mathrm{CH}_3\left(\mathrm{CH}_2\right)_{14} \mathrm{COOH} \\$

Acid palmitic acid has higher melting point, because it has two more methylene groups.

Giving it a greater surface area and therefore more intermolecular van der waals interact than the myristic acid.

stearic arid $\mathrm{CH}_3\left(\mathrm{CH}_2\right)_{16} \mathrm{COOH}$

linoleic acid $\mathrm{C}_{18} \mathrm{H}_{32} \mathrm{O}_2$ (two double bond)

Stearic acid has higher Melting point, because it does not have any Carbon-Carbon double bonds, whereas linoleic acid has two cis double bonds which prevent the molecules from packing closely together.

Oleic Acid and Linoleic acid.

$\mathrm{C}_{18} \mathrm{H}_{34} \mathrm{O}_2$ -one double bond (cis)

Acid palmitic acid has higher melting point, because it has two more methylene groups.

For more such question on methylene group.

brainly.com/question/4279223

#SPJ4

4 0

1 year ago

Complete the following reaction :<br><br>HC≡CH + HBr→?

Morgarella [4.7K]

hope it helps you .............

3 0

2 years ago

At a certain temperature, 0.760 mol SO3 is placed in a 1.50 L container. 2SO3(g) = 2SO2(g) + O2(g)At equilibrium, 0.130 mol O2 i

olganol [36]

Answer:

$K_c=0.0867$

Explanation:

Moles of SO₃ = 0.760 mol

Volume = 1.50 L

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.760}{1.50\ L}$

[SO₃] = 0.5067 M

Considering the ICE table for the equilibrium as:

$\begin{matrix} & 2SO_3_{(g)} & \rightleftharpoons & 2SO_2_{(g)} & + & O_2_{(g)}\\At\ time, t = 0 & 0.5067 &&0&&0 \\ At\ time, t=t_{eq} & -2x &&2x&&x \\----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:- &0.5067-2x&&2x&&x\end{matrix}$

Given:

Equilibrium concentration of O₂ = 0.130 mol

Volume = 1.50 L

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.130}{1.50\ L}$

[O₂] = x = 0.0867 M

[SO₂] = 2x = 0.1733 M

[SO₃] = 0.5067-2x = 0.3334 M

The expression for the equilibrium constant is:

$K_c=\frac {[SO_2]^2[O_2]}{[SO_3]^2}$

$K_c=\frac{(0.3334)^2\times 0.0867}{(0.3334)^2}$

$K_c=0.0867$

5 0

4 years ago

Ydrogen and fluorine combine according to the equation h2(g) + f2(g) → 2 hf(g) if 5.00 g of hydrogen gas are combined with 38.0

Alexxx [7]

<span> Molar mass (H2)=2*1.0=2.0 g/mol
Molar mass (F2)=2*19.0=38.0 g/mol
Molar mass (HF)=1.0+19.0=20.0 g/mol

5.00 g H2 * 1mol H2 /2 g H2=2.50 mol H2
38.0 g F2*1mol F2/38.0 g F2=1.00 mol F2

H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 1 mol
From problem 2.50 mol 1 .00mol

We can see that excess of H2, and that F2 is a limiting reactant.
So, the amount of HF is limited by the amount of F2.

</span> H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 2 mol
From problem 1.00 mol 2.00mol

2.00 mol HF can be formed.

2.00 mol HF*20.0g HF/1mol HF=40.0 g HF can be formed

4 0

3 years ago