Answer:
a molecule with two of the same element
Explanation:
Answer:
Explanation:
Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²
2x 0.007 50 + x
![K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BAg%24%5E%7B%2B%7D%24%5D%24%5E%7B2%7D%24%5BCO%24_%7B3%7D%5E%7B2-%7D%24%5D%7D%20%3D%20%282x%29%5E%7B2%7D%5Ctimes%200.00750%20%3D%208.10%20%5Ctimes%2010%5E%7B-12%7D%5C%5C0.0300x%5E%7B2%7D%20%3D%208.10%20%5Ctimes%2010%5E%7B-12%7D%5C%5Cx%5E%7B2%7D%20%3D%202.70%20%5Ctimes%2010%5E%7B-10%7D%5C%5Cx%20%3D%20%5Csqrt%7B2.70%20%5Ctimes%2010%5E%7B-10%7D%7D%20%3D%20%5Cmathbf%7B1.64%5Ctimes%2010%5E%7B5%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20maximum%20concentration%20of%20Ag%24%5E%7B%2B%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B1.64%5Ctimes%2010%5E%7B-5%7D%7D%5Ctextbf%7B%20mol%2FL%20%7D%7D%24%7D)
Answer: 4.15234 m
512 g H2O * 
= 0.512 kg H2O
Nitric Acid: HNO3 = 1.008 + 14.007 + 3(15.999) = 63.012 g/mol
H = 1.008 g/mol
N = 14.007 g/mol
O3 = 3*15.999
134 g HNO₃ * 
= 2.126 mol
m = 
= 4.15234 m
<span>write out the balance equation
3NaOh+H3PO4->Na3PO4+3H2O
You are given everything needed to calculate
q=heat transfer=2.2*10^2,
H3PO4 moles= 1.5*10^-3,
NaOH moles=5.0*10^-3
equation is deltaHneutraliztion=q/Moles of limiting reagent
H3PO4 is limiting reagent because lowest moles, and is used up first
Now plug in variables
DeltaH=2.2*10^2(1.5*10^3)= 146.67kj/mole
Notice we had to convert J to kj,</span>
