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olasank [31]
2 years ago
12

A gas has a pressure of 3 atm at 350°C. What will its pressure be at 250°C? The volume and amount of gas is constant. Hint: Conv

ert temperature to Kelvin before calculating (°C + 273 = K)
\
Chemistry
1 answer:
vichka [17]2 years ago
7 0

<u>We are given:</u>

P1 = 3 atm                  T1 = 623 K <em>(350 + 273)</em>

P2 = x atm                 T2 = 523 K <em>(250 + 273)</em>

<em />

<u>Solving for x:</u>

From the idea gas equation:

PV = nRT

since number of moles (n) , Volume (V) and the Universal Gas constant(R) are constants;

P / T = k   (where k is a constant)

the value of  k will be the same for a gas with variable pressure and temperature and constant moles and volume

Hence, we can say that:

P1 / T1 = P2 / T2

3 / 623 = x / 523

x = 523 * 3 / 623

x = 2.5 atm (approx)

Therefore, the final pressure is 2.5 atm

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B

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Can snow have viscosity?
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Answer: Yes

Explanation: It can because snow is wet and anything that is wet can

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Question 2*
Paraphin [41]

Answer:

Sc(OH)₃ = 96 g/mol

Explanation:

Gram formula mass:

Gram formula mass is the atomic mass of one mole of any substance.

It can be calculated by adding the mass of each atoms present in substance.

Sc(OH)₃:

Atomic mass of Sc = 45 amu

Atomic mas of O = 16 amu

There are three atoms of O = 16 ×3

Atomic mas of H = 1 amu

There are three atoms of H = 1 ×3

Gram formula mass of Sc(OH)₃:

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6 0
3 years ago
Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10<br> DONE
nadezda [96]

Answer:

The pH of the solution is 5.31.

Explanation:

Let "\alpha is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

                  HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}

Initial                  C                         0            0

Equilibrium        c(1- \alpha)              c\alpha c\alpha

Dissociation constant = Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}

=\frac{c\alpha^{2}}{(1-\alpha)}

In this case weak acids \alpha is very small so, (1-\alpha ) is taken as 1.

Ka=C\alpha^{2}

\alpha=\sqrt\frac{ka}{c}

From the given the concentration = 0.050 M

Substitute the given value.

\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}

[H_{3}O^{+}]=c\alpha

[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}

pH= -log[H_{3}O^{+}]

=-log[4.9\times10^{-6}]

=6-log 4.9= 5.31

Therefore, The pH of the solution is 5.31.

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