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dimaraw [331]
2 years ago
10

A geochemist in the field takes a small sample of the crystals of mineral compound X from a rock pool lined with more crystals o

f X. He notes the temperature of the pool, 26.° C, and caps the sample carefully. Back in the lab, the geochemist dissolves the crystals in 3.00 L of distilled water. He then filters this solution and evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.36 kg1) Using only the information above can you calculate the solubility of X in water at 26 degrees Celsius? yes or no2) If yes calculate the solubility. Round answer to 2 signifacnt digits
Chemistry
1 answer:
Anna [14]2 years ago
3 0

Answer:

The solubility is 0.13 g/mL

Explanation:

Step 1: Data given

Temperature = 26.0 °C =299 K

Volume = 3.00 L

The mass of the crystals, after washing and drying = 0.36 kg = 360 grams

step 2: Calculate the solubility

3.00 L of water contains 360 grams of crystals

For 1.00L of water we'll have 360 / 3 = 130 grams of crystals

This means we have 130 grams of crystals in 1 L, this gives us a solubility of 130g/L

In 1000 mL we have 130 grams crystal

in 1 mL we have 130/1000 = 0.130 grams of crystals

The solubility is 0.13 g/mL

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Answer:

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2 years ago
0.45 g of hydrated sodium carbonate crystals were heated until 3.87 of anhydrous power remained.
snow_tiger [21]

Formula of hydrated sodium carbonate : Na₂CO₃.10H₂O, so moles of water in one mole of hydrated salt = 10

<h3>Further explanation</h3>

Hydrate is a compound that binds water (H₂O), usually in the form of crystals/ solids

If these compounds are dissolved in water or heated, the hydrates can decompose:

Example: X.YH₂O (s) → X (aq) + YH₂O (l)

The formula for the hydrated compound contains: YH2O

The mole ratio shows the ratio of the coefficients of the hydrate compound

10.45 hydrated sodium carbonate(Na₂CO₃.xH₂O) were heated until 3.87 of 3.87of anhydrous (Na₂CO₃) remained, so

mass H₂O released :

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\tt \dfrac{3.87}{105,9888}\div \dfrac{6.58}{18}=0.0365\div 0.3655=1\div 10

6 0
2 years ago
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V - volume - 0.600 m³
n - number of moles
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T - temperature - 282 K
substituting these values in the equation 
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to find the mass of gas
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