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1 year ago

Some thermometers contain alcohol. Alcohol is used in thermometers because it:.

1 answer:

6 0

Because alcohol expands over small temp changes, hence while a vial of it is used in sprinklers as the stop plug; the room gets hotter, the plug shatters.

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Question 5 of 5

rjkz [21]

Probably Fresh vegetables, it can rot out, that’d be my guess, it’s not canned

8 0

2 years ago

Read 2 more answers

When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of

Yanka [14]

Answer:

Mass of carbon dioxide produced = 52.8 g

Explanation:

Given data:

Mass of carbon react = 14.4 g

Mass of oxygen = 56.5 g

Mass of oxygen left = 18.1 g

Mass of carbon dioxide produced = ?

Solution:

C + O₂ → CO₂

Number of moles of C:

Number of moles = mass/ molar mass

Number of moles = 14.4 g/ 12 g/mol

Number of moles = 1.2 mol

18.1 g of oxygen left it means carbon is limiting reactant.

Now we will compare the moles of C with CO₂.

C : CO₂

1 : 1

1.2 : 1.2

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 1.2 mol × 44 g/mol

Mass = 52.8 g

8 0

2 years ago

You know that 8.9 moles of solute particles are dissolved in that liquid solution. If the molarity of the solution is 25 M, how

pychu [463]

Answer:

V = 0.356 L

Explanation:

In this case, we need to use the following expression:

M = n/V (1)

Where:

M: molarity of solution (mol/L or M)

n: moles of solute (moles)

V: Volume of solution (Liters)

From these expression, we can solve for V:

V = n/M (2)

Now, replacing the given data we can solve V:

V = 8.9 / 25

V = 0.356 L

5 0

3 years ago

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

Initial concentration of $\rm HA$ without any dissociation:

$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .