The unites of measurement in a data table should be shown in the headings of some columns.
The balanced chemical equation that represents the reaction is as follows:
<span>SrBr2(aq) + 2AgNO3(aq) → Sr(NO3)2(aq) + 2AgBr(s)
</span>
From the periodic table:
mass of silver = 108 grams
mass of bromine = 80 grams
molar mass of silver bromide = 108 + 80 = 188 grams
number of moles = mass / molar mass
number of moles of produced precipitate = 3.491/188 = 0.018 moles
From the balanced equation:
1 mole of strontium bromide produces 2 moles of silver bromide. Therefore, to calculate the number of moles of <span>strontium bromide that produces 0.018 moles of silver bromide, you will just do a cross multiplication as follows:
amount of </span><span>strontium bromide = (0.018x1) / 2 = 9.28 x 10^-3 moles</span>
About 2,500 grams Ans balkfdoaks;
Answer:
Explanation:Since the compound X has no net-dipole moment so we can ascertain that this compound is not associated with any polarity.
hence the compound must be overall non-polar. The net dipole moment of compound is zero means that the vector sum of individual dipoles are zero and hence the two individual bond dipoles associated with C-Cl bond must be oriented in the opposite directions with respect to each other.]
So we can propose that compound X must be trans alkene as only in trans compounds the individual bond dipoles cancel each other.
If one isomer of the alkene is trans then the other two isomers may be cis .
Since the two alkenes give the same molecular formula on hydrogenation which means they are quite similar and only slightly different.
The two possibility of cis structures are possible:
in the first way it is possible the one carbon has two chlorine substituents and the carbon has two hydrogens.
Or the other way could be that two chlorine atoms are present on the two carbon atoms in cis manner that is on the same side and two hydrogens are also present on the different carbon atoms in the same manner.
Kindly refer the attachments for the structure of compounds:
Answer:
1.98 M
Explanation:
Given data
- Initial volume (V₁): 93.2 mL
- Initial concentration (C₁): 2.03 M
- Volume of water added: 3.92 L
Step 1: Convert V₁ to liters
We will use the relationship 1 L = 1000 mL.

Step 2: Calculate the final volume (V₂)
The final volume is the sum of the initial volume and the volume of water.

Step 3: Calculate the final concentration (C₂)
We will use the dilution rule.
