It is colorless and oderless is a physical property by telling what color.
Answer : The solubility of this compound in g/L is 
.
Solution : Given,
Molar mass of 
= 114.945g/mole
The balanced equilibrium reaction is,
At equilibrium s s
The expression for solubility constant is,
![K_{sp}=[Mn^{2+}][CO^{2-}_3]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BMn%5E%7B2%2B%7D%5D%5BCO%5E%7B2-%7D_3%5D)
Now put the given values in this expression, we get
The value of 's' is the molar concentration of manganese ion and carbonate ion.
Now we have to calculate the solubility in terms of g/L multiplying by the Molar mass of the given compound.
Therefore, the solubility of this compound in g/L is .
Allene (1,2-propadiene) has point group D2d, itself is achiral because it has two planes of symmetry. ... An allene with substituents on one terminal carbon atom are unlike and substituent on other terminal carbon atoms are same, allene will be achiral. It will have one symmetry plane.
Hope this helped :)
Answer:
Mass percent N₂ = 89%
Mass percent H₂ = 11%
Explanation:
First we <u>use PV=nRT to calculate n</u>, which is the total number of moles of nitrogen and hydrogen:
- 1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K
So now we know that
- MolH₂ + MolN₂ = 0.307 mol
and
- MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g
So we have a <u>system of two equations and two unknowns</u>. We use algebra to solve it:
Express MolH₂ in terms of MolN₂:
- MolH₂ + MolN₂ = 0.307 mol
Replace that value in the second equation:
- MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49
- (0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49
- 0.614 - 2MolN₂ + 28molN₂ = 3.49
Now we calculate MolH₂:
- MolH₂ + MolN₂ = 0.307 mol
Finally, we convert each of those mol numbers to mass, to <u>calculate the mass percent</u>:
- N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂
- H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂
Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%
Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%
ANSWER:A.blood
WHY: because it make the most sense and it seem right
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