The compound that will have a sweet smell would be the one, whereby the molecular formula closely resembles that of an ether
R-O-R.
I believe the third one
Answer:
Explanation:
From the given information:
The concentration of metal ions are:
![[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%3D%20%5Cdfrac%7B0.003474%20%5C%20M%20%5Ctimes%2020.49%20%5C%20mL%7D%7B10.0%20%5C%20mL%7D)
![[Ca^{2+}]=0.007118 \ M](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%3D0.007118%20%5C%20M)
![[Mg^2+] = \dfrac{0.003474 \ M\times (26.23 - 20.49 )mL}{10.0 \ mL}](https://tex.z-dn.net/?f=%5BMg%5E2%2B%5D%20%3D%20%5Cdfrac%7B0.003474%20%5C%20M%5Ctimes%20%2826.23%20%20-%2020.49%20%29mL%7D%7B10.0%20%5C%20mL%7D)

Mass of Ca²⁺ in 2.00 L urine sample is:

= 0.1598 g
Mass of Ca²⁺ = 159.0 mg
Mass of Mg²⁺ in 2.00 L urine sample is:

= 0.3461 g
Mass of Mg²⁺ = 346.1 mg
228 grams
start with mass of Cr multiply by molar mass of Cr mole to mole ratio between Cr and Cr2O3 times molar mass of Cr2O3
Answer:
0.4 M
Explanation:
The process that takes place in an aqueous K₂HPO₄ solution is:
First we <u>calculate how many K₂HPO₄ moles are there in 200 mL of a 0.2 M solution</u>:
- 200 mL * 0.2 M = 40 mmol K₂HPO₄
Then we <u>convert K₂HPO₄ moles into K⁺ moles</u>, using the <em>stoichiometric coefficients</em> of the reaction above:
- 40 mmol K₂HPO₄ *
= 80 mmol K⁺
Finally we <em>divide the number of K⁺ moles by the volume</em>, to <u>calculate the molarity</u>:
- 80 mmol K⁺ / 200 mL = 0.4 M