List of the metals that form only one type of ion (that is, metals whose charge is invariant from one compound to another) is below,
Here , we have to list the metals that form only one type of ion (that is, metals whose charge is invariant from one compound to another) and have to identify the group numbers of these metals.
- Hydrogen H+
- Lithium Li+
- Sodium Na+
- Potassium K+
- Rubidium Rb+
- Cesium Cs+
- Beryllium Be2+
- Magnesium Mg2+
- Calcium Ca2+
- Strontium Sr2+
- Barium Ba2+
- Aluminum Al3+
- Zinc Zn2+
- Scandium Sc3+
- Silver Ag+
Co is stable in both +2 and +3 oxidation state. So it can variable. Li only have +1 charge. So it is invariable from one compound to other.
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I'm going to assume that you mean't mole instead of mile. One mole of O2 has approximately the same mass of one "mole" of N2. I say this because if these elements were rounded by their atomic mass, N would stay as 14, O would round up to 16, and we wouldn't; in this case, O2 would have approximately the same mass as F because F rounds up to 19 and the different between O and N is only 2 while the difference between O and F is 3.
Answer:
2Ag + CaCl2
Explanation:
https://en.intl.chemicalaid.com/tools/equationbalancer.php?equation=Ca+%2B+AgCl+%3D+Ag+%2B+CaCl2 use that website for chemistry
From a stock solution of 3.00 m nitric acid, 9.391 ml of stock solution is needed to create a 0.161 m nitric acid solution, which has a total volume of 175 ml of the diluted solution.
A chemical reagent is present in vast quantities as a stock solution. It has a uniform concentration. Examples of typical stock solutions in laboratories are nitric acid and hydrochloric acid. These play a critical role in creating the titration-related solution preparations.
We know the formula for dilution type problems
M1 VI = M2 V 2
Where,
M, = initial molarity
V , = initial Volume
M2 = final molarity
V 2 = final Volume
Hene given -
M, = 3.00 M
VI = ?
M2 = 0.161M
V 2 = 175 ml
Accordingly ' MI V1 = M2 V 2
V1 =
V1= (0.161M*175ml)/ 3.00M
v1 = 9.391
The required volume of Stock solution is 9.391ml.
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Answer:
See explanation and image attached
Explanation:
When an alkene is reacted with cold, dilute and alkaline KMnO4, the product of the reaction is a 1,2-diol (vicinal diol). The products exhibit a “syn” stereochemistry with the addition of the two -OH groups to the same side of the alkene.
The reaction of cyclopentene with cold, dilute, alkaline KMnO4 is shown in the image attached.