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Basile [38]
3 years ago
5

Empirical formula for C2H6O4

Chemistry
1 answer:
Charra [1.4K]3 years ago
8 0

Empirical formula would be CH3O2. This is the multiple that can be multiplied by any number to get a molecule.

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Is baking soda is dissolved in vinegar and bubbles appear a chemical change?
nydimaria [60]
Yes baking soda is dissolved in vinegar and bubbles appear is a CHEMICAL CHANGE .
8 0
3 years ago
Read 2 more answers
Mass=35g, density=5g/cm3. what is the volume?
nevsk [136]

Answer:

7 cm^3

Explanation:

Density = Mass/Volume

5 = 35/x

multiply both sides by x to get

5x = 35

then divide by 5

so x = 7 cm^3

8 0
2 years ago
You chill your reaction in an ice bath when adding nitric acid to the sulfuric acid. why?
Andrej [43]
The reaction of nitric acid and sulfuric acid is highly exothermic so it releases a lot of heat. If the temperature is not controlled, the reaction could go into thermal runaway, which is potentially extremely hazardous.
4 0
3 years ago
A hypothesis is made before the experiment is conducted. <br> True or fasle
Viefleur [7K]

It should be at beginning. A hypothesis is called an educated guess of what might happen in the experiment.

Please brainliest

5 0
3 years ago
If you are given the molarity of a solution, what additional information would you need to find the weight/weight percent (w/w%)
Ludmilka [50]

Answer:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

Explanation:

Hello,

In this case, since molarity is mathematically defined as the moles of solute divided by the volume of solution and the weight/weight percent as the mass of solute divided by the mass of solution, we need:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

For instance, if a 1-M solution of HCl has a density of 1.125 g/mL, we can compute the w/w% as follows:

w/w\%=1\frac{molHCl}{L\ sln}*\frac{36.45gHCl}{1molHCl}*\frac{1L\ sln}{1000mL\ sln}*\frac{1mL\ sln}{1.125g\ sln}    *100\%\\\\w/w\%=3.15\%

Whereas the first factor corresponds to the molar mass of HCl, the second one the conversion from L to mL of solution and the third one the density to express in terms of grams of solution.

Regards.

4 0
3 years ago
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