The given question is incomplete. The complete question is :
A chemist prepares a solution of barium acetate by measuring out 32 g of barium acetate into a 350 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium acetate solution. Round your answer to significant digits.
Answer: The concentration of barium acetate solution is 0.375 mol/L
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in ml
moles of =
Now put all the given values in the formula of molality, we get
Therefore, the concentration of solution is 0.375 mol/L
Except D.
Pollution doesn't prevent cancer.
Unless the pollution is made of a cancer-curing medicine.
Answer:
The electronic configuration that are incorrectly written is 1s²2s³2p⁶, 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴.
Explanation:
The electronic configuration of the elements corresponds to how all the electrons of an element are arranged in energy levels and sub-levels.
There are 7 energy levels —from 1 to 7— whose sublevels are described as s, p, d and f.
All electronic configurations begin with the term "1s" —corresponding to the sublevel s of level 1— so 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴ are incorrectly written. In addition, 4s²3d¹⁰4p⁷ is written incorrectly because is impossible to jump from the sublevel "s" to the sublevel "d" —which is found from level 3 and up— without passing through the sublevel "p".
In the case of 1s²2s³2p⁶, the wrong thing is that the sublevel "s" can only hold two electrons, not three.
The other options are correctly written.
Answer:
Option b. Decomposition
Followed by a reduction process using charcoal
Explanation:
Lead can be obtained from lead nitrate by thermal decomposition of lead nitrate as shown below:
2Pb(NO3)2 —> 2PbO + 4NO2 + O2
The PbO obtained is reduced by charcoal(C) to obtain the metallic Pb as shown below:
2PbO + C —> Pb + CO2
Answer:
0.0125 moles of HCl
Explanation:
(0.50 mol/L)(0.025 L) = 0.0125 moles of HCl