The molar concentration of the KI_3 solution is 0.0833 mol/L.
<em>Step 1</em>. Calculate the <em>moles of S_2O_3^(2-)</em>
Moles of S_2O_3^(2-) = 25.00 mL S_2O_3^(2-) ×[0.200 mmol S_2O_3^(2-)/(1 mL S_2O_3^(2-)] = 5.000 mmol S_2O_3^(2-)
<em>Step 2</em>. Calculate the <em>moles of I_3^(-)
</em>
Moles of I_3^(-) = 5.000 mmol S_2O_3^(2-)))) × [1 mmol I_3^(-)/(2 mmol S_2O_3^(2-)] = 2.500 mmol I_3^(-)
<em>Step 3</em>. Calculate the <em>molar concentration of the I_3^(-)</em>
<em>c</em> = "moles"/"litres" = 2.500 mmol/30.00 mL = 0.083 33 mol/L
Answer:
The compound with the correct formula is;
A. MNO₃
Explanation:
The number of oxidation states in the metal, M = One oxidation state
The formula of the compound formed by the metal, M = MHCO₃
We note that the ion HCO₃⁻, known as hydrogen carbonate has an oxidation number of -1
Similarly nitrate, NO₃⁻ has an oxidation number of -1, therefore, the metal M can form similar compound formed with HCO₃⁻ with nitrate, and we have;
The possible compounds formed by the metal 'M' includes MHCO₃ and MNO₃.
Answer:
The mass of radon that decompose = 63. 4 g
Explanation:
R.R = P.E/(2ᵇ/ⁿ)
Where R.R = radioactive remain, P.E = parent element, b = Time, n = half life.
Where P.E = 100 g , b = 5.55 days, n = 3.823 days.
∴ R.R = 100/
R.R = 100/
R.R = 100/2.73
R.R = 36.63 g.
The mass of radon that decompose = Initial mass of radon - Remaining mass of radon after radioactivity.
Mass of radon that decompose = 100 - 36.63
= 63.37 ≈ 63.4 g
The mass of radon that decompose = 63. 4 g
Nacl aq. is the best conductor of electricity as in aques state the ions become lose or mobile which are free to move and conduct electricity.
