Moles = mass/molar mass
moles = 2.3
molar mass = 278
=> mass = moles*molar mass = 639.4g
The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]
<h3>What is Balanced Chemical Equation ?</h3>
The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
The equation for aqueous solution of H₂CO₃ is
H₂CO₃ → H₂O + CO₂
The charge balance equation is
[HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]
Thus from the above conclusion we can say that The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]
Learn more about the Balanced Chemical equation here: brainly.com/question/26694427
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948 or 9.48 x 10^2
There are two sets of rules for significant figures
• One set for addition and subtraction
• Another set for multiplication and division
You used the set for multiplication and division.
This problem involves addition and subtraction, and the rule is
The number of places after the decimal point in the answer must be <em>no greater than the number of decimal places in every term</em> in the sum.
Thus, we have
78.9
+890.43
-21.
= 948.33
The "21" term has the fewest digits after the decimal point (none), so the answer must have no digits after the decimal point.
To the correct answer is 948 = 9.48 x 10^2. It has three significant figures.
Answer:
most likely that (2) the replicated experiment was performed incorrectly.
Why, u ask? u dare question me:
1- The initial experiment invalidness cannot be proven.
2- <em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>s</u></em><em><u>e</u></em><em><u>c</u></em><em><u>o</u></em><em><u>n</u></em><em><u>d</u></em><em><u> </u></em><em><u>a</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em><u>r</u></em><em><u> </u></em><em><u>i</u></em><em><u>s</u></em><em><u> </u></em><em><u>c</u></em><em><u>o</u></em><em><u>r</u></em><em><u>r</u></em><em><u>e</u></em><em><u>c</u></em><em><u>t</u></em>
3- Different labaratories does not effect the outcome, as long as the parameter and environment of the replicated experiment is the same as when the initial experiment was conducted.
4- Already knowing the data and errors would increase the precision of the replicated experiment.
5- Change in variables should still be in the objective (or purpose) of the experiment, thus, major difference in the outcome should not happen.
happy learning!
