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lbvjy [14]
3 years ago
8

Help help help help​

Chemistry
1 answer:
Art [367]3 years ago
6 0

Answer:

1) saturated

2) unsaturated

3) saturated

4) unsaturated

5) saturated

6) saturated

7) saturated

8) unsaturated

9) saturated

10) saturated

40g of saturated strawberry powder in 100 ml of milk is more soluble.

Explanation:

We define a saturated solution as a solution that holds just as much solute as it can normally hold at a given temperature. Hence, a saturated solution. is unable to dissolve more solute and undissolved solutes begins to appear or gasses are given off.

An unsaturated solute contains less solute than it can normally hold at that particular temperature hence it can still dissolve more solute.

In the closure, we must note that the greater the volume of solvent, the greater the solubility of the solute. Hence it follows that; 40g of saturated strawberry powder in 100 ml of milk is more soluble.

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What is the mass in grams of 9.76 × 1012 atoms of naturally occurring beryllium?
Ksivusya [100]

The atomic mass of beryllium (Be) is 9 g/mole

Now, 1 mole of any substance contains avogadro's number of that substance. Therefore:

1 mole of Be contains 6.023  10^23 atoms of Be

In other words:

9 grams of Be contains 6.023*10^23 atoms of Be

Therefore, the mass corresponding to 9.76 * 10^12 atoms of Be is:

= 9 g * 9.76 10^12 atoms/6.023*10^23 atoms

= 1.458 * 10^-10 g (or) 1.46 * 10^-10 g


8 0
3 years ago
Read 2 more answers
er solutions can be produced by mixing a weak acid with its conjugate base or by mixing a weak base with its conjugate acid. The
liubo4ka [24]

Answer:

You need to add 19,5 mmol of acetates

Explanation:

Using the Henderson-Hasselbalch equation:

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For the buffer of acetates:

pH = pKa + log₁₀ [CH₃COO⁻]/[CH₃COOH]

As pH you want is 5,03, pka is 4,74 and milimoles of acetic acid are 10:

5,03 = 4,74 + log₁₀ [CH₃COO⁻]/[10]

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Thus, to produce an acetate buffer of 5,03 having 10 mmol of acetic acid, you need to add 19,5 mmol of acetates.

I hope it helps!

7 0
4 years ago
A mole of ethyl alcohol weighs 46 g. how many grams of ethyl alcohol are needed to produce 1 l of a 2-millimolar (2 mm) solution
Thepotemich [5.8K]
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Ethyl alcohol molecular weight is 46g/mole, then 2 mili mole of glucose is= 2x 10^-3 mole x 46 grams/mole= 92 grams x 10^-3= 92mg
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3 0
3 years ago
I am a Non reactive, non metal, with 8 Valence electrons. What element am I?
devlian [24]

Answer:

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Explanation:

5 0
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