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Maslowich
3 years ago
7

How much of a 0.250 M sucrose solution must be used to prepare 400.0 mL of a 0.0310 M solution

Chemistry
2 answers:
Svetach [21]3 years ago
5 0
The answer would be 49.6 mL
kupik [55]3 years ago
3 0

The answer would be 49.6 mL

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PLEASE HELP MEEE!!!!
Assoli18 [71]
It would be 35.8 Calories or calories. Not sure about that part. Hope this helps though.
3 0
3 years ago
Scientific explanations are not always based on empirical evidence <br><br> True or false?
tigry1 [53]
True some explanations are not always based on empirical evidence
4 0
3 years ago
Which type of reactions form salts?
s344n2d4d5 [400]

Neutralization reactions are the reactions type which form salts.

Explanation:

Salts are formed by ionic bonds when the oxidation states of anions and cations are equal and have opposite signs. So one should be highly electronegative in nature and another should be highly electropositive in nature. So the electropositive element will be ready to give electrons and the electronegative element will be ready to accept all the electrons given by the electropositive element. As a whole the compound will be neutrally charged by adding of equal number of positively charged and negatively charged ions.

The reduction or addition of electrons will be occurring in cations and the oxidation or removal of electrons will be occurring in anions.

So the salt formation is based on neutralization reactions.

8 0
3 years ago
Which colors form part of the color wheel?
kicyunya [14]

Answer: primary colors are those you can't get by mix in other colors. They are red, blue, and yellow. secondary colors come from mixing two primary colors.

6 0
3 years ago
How many grams are in 10.25 moles of zinc chromate?
Kobotan [32]

Answer:

1859.4 g of ZnCrO₄ in 10.25 moles

Explanation:

First of all, we determine the molecular formula of the compound:

Zinc → Zn²⁺  (cation)

Chromate → CrO₄⁻²  (anion)

Zinc chromate → ZnCrO₄

Molar mass for the compound is:

Molar mass of Zn + Molar mass of Cr + (Molar mass of O) . 4 = 181.41 g/mol

65.41 g/mol + 52 g/mol + 16 g/mol . 4 = 181.41 g/mol

Let's apply this conversion factor: 10.25 mol . 181.41 g/mol = 1859.4 g

5 0
2 years ago
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