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Sauron [17]
3 years ago
5

How do you actually know the chemical formulae for the ionic compound

Chemistry
1 answer:
qaws [65]3 years ago
3 0
If you have to write the chemical formula of a simple, binary ionic compound given the name of the compound, you follow a set of three steps. Let's go through them using magnesium chloride as an example. Write the symbols for the cation and the anion: Mg and Cl. Determine the charge on the cation and anion.
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A certain first-order reaction a → b is 25% complete in 42 min at 25°c. What is the half-life of the reaction?.
Anika [276]

The half-life of the reaction is 101.9 min

<h3>First order reaction </h3>

When a reaction's pace and reactant concentration are inversely correlated, the process is known as a first-order reaction. To put it another way, the reaction rate doubles when the concentration does. One or two reactants can be present in a first-order reaction, as in the case of the decomposition process.

<h3>The half-life of first-order reaction:-</h3>

The amount of time it takes for the initial concentration of the reactant(s) to decrease by half is known as the half-life of a chemical reaction (abbreviated as "t1/2").

<h3 /><h3>Calculation:-</h3>

a→b

25% reacted means 75% remains

t=42 min

Rate constant

k=(2.303/t)(log a/a-x)

k=(2.303/42)(log 100/100-25 )

k=(0.054) (log 100/75)

k=(0.054)(0.1249)

k=0.0068 per min

half life

t1/2=(0.693/k)

     =(0.693/0.0068)

     =101.9 min

Learn more about first order reaction here :-

brainly.com/question/27754430

#SPJ4

5 0
2 years ago
How can large crystals of phthalic acid be grown compared to smaller crystals?
anastassius [24]
Crystals of phthalic acid can be grown  twice as fast compared to the smaller crystals

8 0
4 years ago
Calculate the percentage by mass of lead in pbco3.
nignag [31]
Pb/pbco3*100
207/207+12+48
207/267*100
=77.53%
5 0
4 years ago
Read 2 more answers
A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is
Colt1911 [192]

Answer:

M_{Na^+}=1.36M

M_{Br^-}=1.58M

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

Once we've got the moles we compute the final volume via:

V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L

Thus, the molarity of the sodium atoms turn out into:

M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

Best regards.

7 0
3 years ago
How is polorization corrected in a simple cell​
valkas [14]

Answer:

hydrogen bubbles rise to the surface of the electrolyte and escape into the air, some remain on the surface of the anode. If enough bubbles remain around the anode, the bubbles form a barrier that increases internal resistance.

6 0
3 years ago
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