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adelina 88 [10]
3 years ago
8

Which elements are solid at room temperature

Chemistry
1 answer:
Kryger [21]3 years ago
8 0

Answer:bromine , neon , helium , argon , lithium , beryllium

Explanation:

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I RLLY NEED HELP. Use Image B (picture above) and calculate the density of a ring that has a mass of 32 grams. Read Page 11 "Mea
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Answer:

Mass of ring = 32 g

Volume of ring = 4 mL

Density of ring = 8 g/mL

Explanation:

From the question given above, the following data were obtained:

Mass of ring = 32 g

Volume of water = 64 mL

Volume of water + ring = 68 mL

Density of ring =?

Next, we shall determine the volume of the ring. This can be obtained as follow:

Volume of water = 64 mL

Volume of water + ring = 68 mL

Volume of ring =?

Volume of ring= (Volume of water + ring) – (Volume of water)

Volume of ring = 68 – 64

Volume of ring = 4 mL

Finally, we shall determine the density of the ring. This can be obtained as follow:

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Density = mass / volume

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8 0
3 years ago
Which solution has the same boiling point as 0.25 mol CaCl2 dissolved in 1000 g water?
Rufina [12.5K]
The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality). 

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<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.

5 0
3 years ago
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