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Natali [406]
3 years ago
15

Exactly 5000 mL of air at 223K is warmed and has a new volume of 8.36 liters. What is the new temperature?

Chemistry
1 answer:
34kurt3 years ago
7 0

Answer:

The new temperature is 373 K

Explanation:

Step 1: Data given

Volume air = 5000 mL = 5.0 L

Temperature = 223K

New volume = 8.36 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒V1 = the initial volume = 5.0 L

⇒T1 = the initial temperature = 223 K

⇒V2 = the new volume = 8.36 L

⇒T2 = the new temperature

5.0/223 = 8.36 /T2

T2 = 373 K

The new temperature is 373 K

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Multiply the values.

n = cv

Where

n =mol

c = concentration

v = volume

n = 2.5×1.5 = 3.75mol

3 0
3 years ago
Which of the following chemical symbols represents the element silver?
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The chemical symbol that represents the element silver is B. Ag
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If a carbon atom is bonded to two hydrogen atoms and two carbon atoms
Nataliya [291]

Answer:

That molecule is a propane.

Explanation:

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3 0
3 years ago
Atypicalaspirintabletcontains325mgofacetylsalicylic acid (HC9H7O4). Calculate the pH of a solution that is prepared by dissolvin
Sergio [31]

Answer:

\boxed{2.65}

Explanation:

1. Mass of acetylsalicylic acid (ASA)

m = \text{2 tablets} \times \dfrac{\text{325 mg}}{\text{1 tablet}} = \text{750 mg}

2. Moles of ASA

HC₉H₇O₄ =180.16 g/mol

n = \text{750 mg} \times \dfrac{\text{1 mmol}}{\text{180.16 mg }} = \text{4.163 mmol}

3. Concentration of ASA

c = \dfrac{\text{4.163 mmol}}{\text{237 mL}} = \text{0.01757 mol/L}

4. Set up an ICE table

\begin{array}{ccccccc}\text{HA} & + & \text{H$_{2}$O}& \, \rightleftharpoons \, &\text{H$_{3}$O$^{+}$} & + &\text{A}^{-}\\0.01757 & & & &0 & & 0 \\-x & & & &+x & & +x \\0.01757-x & & & &x & & x \\\end{array}\\

5. Solve for x

K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}]\text{A}^{-}]} {\text{[HA]}} = 3.33 \times 10^{-4}\\\\\dfrac{x^{2}}{0.01757 - x} = 3.33 \times 10^{-4}\\\\\textbf{Check that }\mathbf{x \ll 0.01757}\\\\\dfrac{ 0.01757 }{3.33 \times 10^{-4}} = 53 < 400\\\\\text{The ratio is less than 400. We must solve a quadratic equation.}\\\\x^{2} = 3.33 \times 10^{-4}(0.01757 - x) \\\\x^{2} = 5.851 \times 10^{-6} - 3.33 \times 10^{-4}x\\\\x^{2} + 3.33 \times 10^{-4}x - 5.851 \times 10^{-6} = 0

6. Solve the quadratic equation.

a = 1; b = 3.33 \times 10^{-4}; c = -5.851 \times 10^{-6}

x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{Substituting values into the formula, we get}\\x = 0.002258\qquad x = -0.002591\\\text{We reject the negative value, so}\\x = 0.002258

7. Calculate the pH

\rm [H_{3}O^{+}]= x \, mol \cdot L^{-1} = 0.002258 \, mol \cdot L^{-1}\\\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.002258} = \mathbf{2.65}\\\text{The pH of the solution is } \boxed{\textbf{2.65}}

4 0
3 years ago
Why is amber a great way to see fossils?
horrorfan [7]

Answer:

because the fossils are beutifule

Explanation:

because the fossils are rare in the world

3 0
2 years ago
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