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2 years ago

5

You are asked to determine the identity of an unknown liquid and can measure only one physical property. You see the liquid and

record that's liquid turn into vapor at 100 degrees Celsius. this demonstrates which physical property

1 answer:

Julli [10]2 years ago

5 0

The physical property you measured is the boiling point of the liquid.

Therfore, if the liquid has a boiling point of 100°C, it turns from liquid state to vapor when you heat it up to 100°C.

To conclude, The liquid should be pure water. Because only pure water has a boiling point of 100°C, using this physical property, we can test whatever liquid ans see if it's pure water or not, this test is called the boiling point test.

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Which chemical equation represents the reaction of an Arrhenius acid and an Arrhenius base?

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The chemical equation that represents the reaction of an Arrhenius acid and an Arrhenius base is 1) HC2H3O2 (aq) + NaOH (aq) ---> NaC2H3O2 (aq) + H2O (I)

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3 years ago

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Consider the following reaction:

Katyanochek1 [597]

Answer:

The three statements are true

Explanation:

For the reaction:

I₂O₅(s) + 5CO(g) → I₂(s) + 5CO₂(g)

State oxidation of iodine in I₂O₅ is:

5 O²⁻ = 10⁻

As you have 2 I and the molecule has no charge, <em>oxidation state of I is +5</em>.

The carbon in CO has an oxidation state of +2 and in CO₂ is +4. That means <em>the carbon is oxidized</em>

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An oxidizing agent is a substance that produce the oxidation of the agent that reacts with this one. CO is oxidized because of I₂O₅ is producing its oxidation being <em>the oxidizing agent</em>

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3 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

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3 years ago

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