Answer:
(a) the mass of the water is 3704 g
(b) the mass of the water is 199, 285.7 g
Explanation:
Given;
Quantity of heat, H= 8.37 x 10⁶ J
Part (a) mass of water (as sweat) need to evaporate to cool that person off
Latent heat of vaporization of water, Lvap. = 2.26 x 10⁶ J/kg
H = m x Lvap.
mass in gram ⇒ 3.704 kg x 1000g = 3704 g
Part (b) quantity of water raised from 25.0 °C to 35.0 °C by 8.37 x 10⁶ J
specific heat capacity of water, C, 4200 J/kg.°C
H = mcΔθ
where;
Δθ is the change in temperature = 35 - 25 = 10°C
mass in gram ⇒ 199.2857 kg x 1000 g = 199285.7 g
To ensure that the 9-Fluorenone was totally dry, it had to be washed with methylene chloride. To make sure that methylene chloride is present in a pure solution, sodium sulfate binds to water and precipitates.
<h3>
What is the purpose of the sodium sulfate?</h3>
- Although it has numerous additional uses, sodium sulfate is primarily employed in the production of detergents and in the Kraft process of paper pulping.
- The decahydrate's natural mineral form, mirabilite, accounts for about half of the world's output, with the other half coming from chemical byproducts. Sodium sulfate was used as a drying, isolating, and anhydrous salt for the 9-fluorenone.
- To make sure that methylene chloride is present in a pure solution, sodium sulfate binds to water and precipitates.
- The sodium salt of sulfuric acid is known as sodium sulfate. Na2SO4 is the chemical formula for sodium sulfate. The mineral thenardite, which is also known as anhydrous sulfate, is described as a white, crystalline solid, whereas the decahydrate Na2SO4. 10H2O is also known as Glauber's salt or the mirabilis salt.
To know more about sodium sulfate, refer:
brainly.com/question/23509646
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Answer:
= 20.82 g of BaCl2
Explanation:
Given,
Volume = 200 mL
Molarity = 0.500 M
Therefore;
Moles = molarity × volume
= 0.2 L × 0.5 M
= 0.1 mole
But; molar mass of BaCl2 is 208.236 g/mole
Therefore; 0.1 mole of BaCl2 will be equivalent to;
= 208.236 g/mol x 0.1 mol
= 20.82 g
Therefore, the mass of BaCl2 in grams required is 20.82 g