Answer:
Neutral atoms can be turned into positively charged ions by removing one or more electrons.
Explanation:
If there is an atom that has 9 protons and 9 electrons, removing an electron from the atom will gain a postive charge.
The reaction between copper II chloride and sodium sulfide as well as lead II nitrate and potassium sulfate both produce precipitates.
The solubility of a substance in water is in accordance with the solubility rules. It is possible that a solid product may be formed when two aqueous solutions are mixed together. That solid product is referred to as a precipitate.
Now, we will consider each reaction individually to decode whether or not a precipitate is possible.
- In the first reaction, we have; CuCl2(aq) + Na2S(aq) ---->CuS(s) + 2NaCl(aq). A precipitate (CuS) is formed.
- In the second reaction, Pb(NO3)2(aq) + 2KNO3(aq) ----> PbSO4(s) + KNO3(aq), a precipitate PbSO4 is formed
- In the third reaction, NH4Br(aq) + NaOH(aq) ----->NH3(g) + NaBr(aq) + H2O(l), a precipitate is not formed here.
Learn more: brainly.com/question/11969651
In total dominance, the phenotype sees only one allele in the genotype. Both alleles in the genotype are seen in the phenotype during codominance. In incomplete dominance, in the phenotype, a mixture of the alleles in the genotype is seen.
At STP condition 1 mol of any ideal gas will have a volume of 22.4L
1.75 mol of F2 x 22.4 L / 1 mol = 39.2 L
Answer: The ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
of a solution is 
M
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
pH or pOH is the measure of acidity or alkalinity of a solution.
According to stoichiometry,
1 mole of gives 1 mole of 
Thus 
moles of gives =
moles of
Putting in the values:
![[H^+][OH^-]=10^{-14}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%5BOH%5E-%5D%3D10%5E%7B-14%7D)
![[0.01][OH^-]=10^{-14}](https://tex.z-dn.net/?f=%5B0.01%5D%5BOH%5E-%5D%3D10%5E%7B-14%7D)
![[OH^-]=10^{-12}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-12%7D)
Thus the of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is M
