B and c...will lose electron(s) in forming an Ion.
P is an Anion
b..Fe. and c...Pb form Cations (+) by losing electrons.
d. Se is an Anion.
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Answer:</h3>
0.35 M
<h3>
Explanation:</h3>
<u>We are given;</u>
- Initial volume as 35.0 mL or 0.035 L
- Initial molarity as 12.0 M
- Final volume is 1.20 L
We are required to determine the final molarity of the solution;
- Dilution involves adding solvent to a solution to make it more dilute which reduces the concentration and increases the solvent while maintaining solute constant.
- Using dilution formula we can determine the final molarity.
M1V1 = M2V2
M2 = M1V1 ÷ V2
= (12.0 M × 0.035 L) ÷ 1.2 L
= 0.35 M
Thus, the final concentration of the solution is 0.35 M
Answer:
A noncompetitive inhibitor can only bind to an enzyme with or without a substrate at several places at a particular point in time
Explanation:
this is because It changes the conformation of an enzyme as well as its active site, which makes the substrate unable to bind to the enzyme effectively so that the efficiency of the enzyme decreases. A noncompetitive inhibitor binds to the enzyme away from the active site, altering/distorting the shape of the enzyme so that even if the substrate can bind, the active site functions less effectively and most of the time also the inhibitor is reversible
Answer:
mass of water in hydrate = 2.37 grams
Explanation:
The mass of the hydrated cobalt (III) chloride is the summation of the salt and the water it contains.
This means that:
Total mass of sample = mass of salt + mass of water
Now, we are given that:
total mass of sample = 5.22 grams
mass of salt = mass of sample after heating = 2.85 grams
Substitute to get the mass of water as follows:
5.22 = mass of water in hydrate + 2.85
mass of water in hydrate = 5.22 - 2.85
mass of water in hydrate = 2.37 grams
Hope this helps :)
Answer:
Explanation:
Hello!
In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:
We plug in the mass of water, temperature change and specific heat to obtain:
Now, this enthalpy of reaction corresponds to the combustion of propyne:
Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:
However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:
Now, we solve for the enthalpy of formation of C3H4 as shown below:
So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):
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