Percentage Weight-in-volume is defined as the <em><u>number of grams of a solute in a 100 ml (milliliters) solution.</u></em>
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<u>Percentage Weight-in-volume</u> can tell us about the <em>degree of concentration of a given solution.</em>
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The solute can be <em>crystalline or non-crystalline in nature.</em>
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The <u>number of grams of glucose</u> present in a <u>5% glucose solution</u> is 5 grams.
- This question is based on a Percentage Weight-in-volume. The formula states that:
a% of a glucose solution =<u> a grams of glucose in a 100 mL solution</u>
Hence, 5% glucose solution = 5 grams of glucose / 100 mL solution
Therefore, the <u>number of grams of glucose</u> present in a <u>5% glucose solution</u> is 5 grams.
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Answer:
The second experiment (reversible path) does more work
Explanation:
Step 1:
A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C
<em>(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm</em>
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Irreversible path: w =-Pex*ΔV
⇒ with Pex = 1.00 atm
⇒ with ΔV = 1.20 L
W = -(1.00 atm) * 1.20 L
W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J
<em>(b) The gas is allowed to expand reversibly and isothermally to the same final volume.</em>
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W = -nRTln(Vfinal/Vinitial)
⇒ with n = the number of moles = 0.200
⇒ with R = gas constant = 8.3145 J/K*mol
⇒ with T = 298 Kelvin
⇒ with Vfinal/Vinitial = 2.40/1.20 = 2
W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)
W = -343.5 J
The second experiment (reversible path) does more work
Higher <span>C. higher than when the source is stationary</span>
Answer:
490
98 for 1 mole, Hence for 5 moles 5 X 98 =490.
Explanation:
Brainliest please?
It’s B. Substitution hope this helps
