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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First convert to moles:
0.5113 g / 17 g/mol = 0.0301 mol
Now create a ratio based on the reaction provided to solve for the unknown:
4 NH3 / -905.4 kJ = 0.0301 mol NH3 / x kJ
x = -6.808 kJ
For both of them, used the balanced equation and it’s mole ratio to convert whatever you need to into moles. See the attacked work.
1) D 5 mols
2) A 0.55 mols
Answer:
15.35 g of (NH₄)₃PO₄
Explanation:
First we need to look at the chemical reaction:
3 NH₃ + H₃PO₄ → (NH₄)₃PO₄
Now we calculate the number of moles of ammonia (NH₃):
number of moles = mass / molecular wight
number of moles = 5.24 / 17 = 0.308 moles of NH₃
Now from the chemical reaction we devise the following reasoning:
if 3 moles of NH₃ are produce 1 mole of (NH₄)₃PO₄
then 0.308 moles of NH₃ are produce X moles of (NH₄)₃PO₄
X = (0.308 × 1) / 3 = 0.103 moles of (NH₄)₃PO₄
mass = number of moles × molecular wight
mass = 0.103 × 149 = 15.35 g of (NH₄)₃PO₄
Answer:
ones that can be mixed together
Explanation:
like water or ethanol
