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2 years ago

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A 20. 0-liter flask contains a mixture of argon at 0. 72 atmosphere and oxygen at 1. 65 atmospheres. What is the total pressure

in the flask? A. 0. 93 atm B. 2. 37 atm C. 8. 44 atm D. 18. 6 atm.

1 answer:

exis [7]2 years ago

5 0

The total pressure of the system is equivalent to the sum of all the pressure of the individual gases. The total pressure of the flask is 2.37 atm.

<h3>What is total pressure?</h3>

According to Dalton's law, the total pressure of the system will be equivalent to the total of the pressures exerted by the individual gases present in the system.

The total pressure of gases is given as,

$\rm P = P_{1} +P_{2}+ P_{3}.....P_{n}$

Given,

The pressure of argon gas = 0.72 atm
The pressure of oxygen = 1.65 atm
Total pressure = P

Substituting values in the above equation:

$\begin{aligned}\rm P &= 0.72 + 1.65\\\\&=2.37\;\rm atm\end{aligned}$

Therefore, 2.37 atm is the total pressure of the flask.

Learn more about total pressure here:

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BARSIC [14]

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<h3>What are moles?</h3>

A mole is defined as 6.02214076 × $10^{23}$ of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

A. The number of C atoms in 0.524 mole of C:

6.02214076 × $10^{23}$ x 0.524 mole

3.155601758 atoms =3.155 atoms

B. The number of $SO_2$ molecules in 9.87 moles of $SO_2$ :

6.02214076 × $10^{23}$ x 9.87

59.4385293 molecules= 59.43 molecules

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1.40 x $10^{22}$ ÷ 6.02214076 × $10^{23}$

0.2324754694 x $10^{-1}$ moles.

0.23 x $10^{-1}$ moles.

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2.30x $10^{24}$ ÷ 6.02214076 × $10^{23}$

3.819239854 moles=3.81 moles

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1 year ago

A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin

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Answer : The mass of the water in two significant figures is, $3.0\times 10^1g$

Explanation :

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$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron metal = $0.45J/g^oC$

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$T_2$ = initial temperature of water = $63.5^oC$

Now put all the given values in the above formula, we get

$32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC$

$m_2=30.16g\approx 3.0\times 10^1g$

Therefore, the mass of the water in two significant figures is, $3.0\times 10^1g$

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2 years ago