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den301095 [7]
3 years ago
6

What does acetic mean​

Chemistry
2 answers:
Llana [10]3 years ago
7 0

Answer:

dym like vinegar or acetic acid

Naddika [18.5K]3 years ago
5 0

Answer:

Acetic

Explanation:

organic chemistry of, pertaining to, or producing vinegar

organic chemistry of or pertaining to acetic acid or its derivatives

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Temperature (°C)
Andreyy89
The answer is the solution saturated than more dissolved with is C
5 0
2 years ago
Compound element ratio of 1:2:6<br> Ba(NO3)2+Na2SO4
Karo-lina-s [1.5K]

1:2:6? Ba(NO3)2 +Na2SO4 -----> BaSO4 + 2NaNO3

Ba(NO3)2

BaSO4

NaNO 3

Na2SO4

5 0
4 years ago
Read 2 more answers
Find the final equilibrium temperature when 15.0 g of milk at 13.0 degrees c is added to 148 g of coffee with a temperature of 8
user100 [1]

According to zeroth law of thermodynamics, when two objects are kept in contact, heat (energy) is transferred from one to the other until they reach the same temperature (are in thermal equilibrium). When the objects are at the same temperature there is no heat transfer.

So, at equilibrium, q_{lost}=q_{gain},  q_{lost}= q_{milk} + q_{coffee}

q=m×c×T, where q = heat energy, m = mass of a substance, c = specific heat (units J/kg∙K), T is temperature

q_{lost}= (15X13X4.19)+(148X88.3X4.19), (15+148)X4.19XT_{final}=(15X13X4.19)+(148X88.3X4.19)

T_{final}= 81.37 ° C

6 0
3 years ago
What is the mole fraction of solute in a 3.19 m aqueous solution?
Bumek [7]
Molality= mol/ Kg

if we assume that we have 1 kg of water, we have 3.19 moles of solute. 

the formula for mole fraction --> mole fraction= mol of solule/ mol of solution

1) if we have 1 kg of water which is same as 1000 grams of water. 

2) we need to convert grams to moles using the molar mass of water 

molar mass of H₂O= (2 x 1.01) + 16.0 = 18.02 g/mol

1000 g (1 mol/ 18.02 grams)= 55.5 mol

3) mole of solution= 55.5 moles + 3.19 moles= 58.7 moles of solution

4) mole fraction= 3.19 / 58.7= 0.0543 
4 0
3 years ago
A compound is found to contain 73.23% xenon name 26.77% oxygen by mass. What is the empirical formula for this compound ?
Luba_88 [7]

The empirical formula is XeO₃.

<u>Explanation:</u>

Assume 100 g of the compound is present. This changes the percents to grams:

Given mass in g:

Xenon = 73.23 g

Oxygen = 26.77 g

We have to convert it to moles.

Xe = 73.23/   131.293 = 0.56 moles

O = 26.77/ 16 = 1.67 moles

Divide by the lowest value, seeking the smallest whole-number ratio:

Xe = 0.56/ 0.56 = 1

O = 1.67/ 0.56 = 2.9 ≈3

So the empirical formula is XeO₃.

6 0
3 years ago
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