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valentina_108 [34]
3 years ago
9

Read the descriptions below of two substances and an experiment on each. Decide whether the result of the experiment tells you t

he substance is a pure substance or a mixture, if you can.
• Sample A is 100 g of a coarse grey powder with a faint unpleasant smell. 15 g of the powder are dissolved in ethanol. 0.5 mg of the resulting black solution is carefully dropped onto a thick sheet of paper laid flat in a tray. After 30 minutes the initial round black stain has spread out and faded in color to a deep purple. Additionally, there is a dark green ring surrounding the inner stain.
• Sample B is 100 ml of a clear liquid. The density of the liquid is measured, and turns out to be 0.77 g/ml. The liquid is then cooled in the refrigerator. At 10.0 °C crystals begin to appear until the liquid is about half crystal, half liquid. After 30 minutes, no more crystals appear, even though the temperature is lowered to 6.3 °C.
Is sample A made from a pure substance or a mixture? If the description of the substance and the outcome of the experiment isn't enough to decide, choose "can't decide.
a. mixture
b. pure substance
c. (can't decide)
Is sample B made from a pure substance or a mixture? If the description of the substance and the outcome of the experiment isn't enough to decide, choose "can't decide."
a. pure substance
b. mixture
c. (can't decide)
Chemistry
1 answer:
inessss [21]3 years ago
4 0

Answer:

Sample A - Mixture

Sample B - (can't decide)

Explanation:

We know a mixture as a sample that is made up of two or more substances. Based on the results from the experiment conducted on sample A, the sample is a mixture. Each colour that appeared on the paper represents one of the components of the mixture.

For Sample B, at a particular sharp temperature, the crystals begin to appear. That temperature at which the first crystal appears is actually the melting point of the solid. We were also told that only half of the clear liquid was crystallized meaning that other substances may still be contained in the remaining liquid. Crystallization is a separation technique that depends on differences in melting points of substances. We can't decide if the sample is pure because we have no further information about what happened to the remaining liquid. That would have told us if the liquid remaining was just the solvent used to dissolve B which could have also been evaporated to leave only the pure sample.

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Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

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ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

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3 years ago
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Explanation:

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<em>That means, 1 mole of hydroxide reacts per mole of acid</em>

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0.0200L ₓ (0.345mol / L) = <em>6.90x10⁻³ moles of HClO₂</em>

To neutralize this acid, you need to add the same number of moles of LiOH, that is 6.90x10⁻³ moles. As the LiOH contains 0.250 moles / L:

6.90x10⁻³ moles ₓ (1L / 0.250mol) = 0.0276L of LiOH =

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Answer:

  • <u>Hey </u><u>mate </u>
  • <u>I </u><u>hope </u><u>it </u><u>helps </u>

Explanation:

<h3>Removing Energy: Removing energy will cause the particles in a liquid to begin locking into place. A. Boiling and Evaporation: Evaporation is the change of a substance from a liquid to a gas. Boiling is the change of a liquid to a vapor, or gas, throughout the liquid.</h3>

<h2>PLZ <u>MARK </u><u>ME </u><u>AS </u><u>BRAIN </u><u>LIST </u><u /></h2>

<u>THANKS </u><u />

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