The ionic equation is as below
Ca^2+(aq) + SO4^2-(aq) ---> CaSO4(s)
EXPLANATION
K2SO4(aq) +Cai2(aq) ---> CaSO4(s) + Ki (aq)
ionic equation
= 2K^+(aq) + SO4^2-(aq) + Ca^2+(aq) + 2i^-(aq) --->CaSO4(s) + 2K^+(aq) +2 i^-(aq)
cancel the spectator ions that is 2k^+ and 2i^-
The net ionic equation is therefore
= Ca^2+(aq) + SO4^2-(aq) ----> CaSO4(s)
Answer:
Explanation:
Hello,
In this case, considering that the by-mass percent of water is:
Given such percent and the mass of the sample, we can find the mass of water in grams in the sample by solving for it as shown below:
Best regards.
Please see the attached photo for the answer. Basically, you break the double bond between the Carbons and add Hydrogens to the carbons to fulfill the octet.
Shout out to my co-worker who helped me solve the problem!
<h3><u>Answer</u>;</h3>
C3H4O
<h3><u>Explanation;</u></h3>
Empirical formula is the simplest formula of a compound;
Molar mass CO2 = 44.01
Mass of CO2 produced = 2.053 g
Mass of carbon in original sample = 12.01/44.01 × 2.053
= 0.5603g
Molar mass H2O = 18
Mass of H in original sample = 2/18 ×0.5601
= 0.0622 g
Thus; original sample contained 0.5603g C and 0.0622g H. The balance of the sample was O
Mass of O = 0.8715 - (0.5603 + 0.0622) = 0.249g
The mole ratio of C:H:O will be;
Moles C = 0.5603/12 = 0.0467
Moles H = 0.0622
Moles O = 0.249/16 = 0.01556
C:H:O = 0.0467:0.0622:0.01556
Divide through by 0.01556:
C:H:O = 3:4:1
Empirical formula is thus C3H4O
<span>~He was working without prior knowledge toward creating a periodic table of elements that had been done in Europe in the 1860s
~ scandal when he got married to his second wife before he divorced his first wife
~he was working with data that was not very precise (technology didn't exist at the time to make accurate measurements (e.g., weights) for the things he was working on.) Many elements had also not been discovered yet, so he had to guess at a lot of things when making his conclusions. </span>
