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2 years ago

Define an arrhenius base and describe properties of bases. use an example to explain how an arrhenius base will behave in water.

1 answer:

4 0

An Arrhenius base is a molecule that when dissolved in water will break down to yield an $OH^-$ or hydroxide in solution.

<h3>What is Arrhenius base?</h3>

An Arrhenius base is a compound that increases the OH− ion concentration in aqueous solution.

An Arrhenius base is a substance that, when dissolved in an aqueous solution, increases the concentration of hydroxide, or $OH^-$ , ions in the solution.

Bases Properties

Arrhenius bases that are soluble in water can conduct electricity.

Bases often have a bitter taste and are found in foods less frequently than acids. Many bases, like soaps, are slippery to the touch.

Bases also change the colour of indicators. Red litmus turns blue in the presence of a base (see figure below), while phenolphthalein turns pink.

Some bases react with metals to produce hydrogen gas.

Acids (pH < 7.0) react with bases (pH > 7.0) to produce a salt and water. When equal moles of an acid and a base are combined, the acid is neutralized by the base. The resulting mixture will have a more neutral pH.

An Arrhenius acid is a substance that dissociates in water to form hydrogen ions or protons. In other words, it increases the number of H+ ions in the water. In contrast, an Arrhenius base dissociates in water to form hydroxide ions, $OH^-.$

Example, sodium hydroxide, is added to an aqueous solution. NaOH dissociates into sodium, $Na^+$ , and hydroxide, $OH^-,$ ions.

Learn more about the Arrhenius bases here:

https://brainly.in/question/8273595

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RoseWind [281]

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3 years ago

At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5

Annette [7]

Answer:

4600s

Explanation:

$2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}$

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

$-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt$

If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.

PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.

$-\frac{d[P(B)]}{P(B)}=k*dt$

$-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt$

Integrating we get:

$\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt$

$-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})$

Clearing for t2:

$\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}$

$ln[P(N_{2}O_{5})]=ln(650)=6.4769$

$ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333$

$t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s$

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3 years ago

How to convert 250 in scientific notation

cricket20 [7]

In scientific notation, a number is less than ten but more than one.

Move the decimal point from 0, 250.000 <- this is the same as 250 to between 2 and 5.

I had to move two spaces.

2.5^2

I hope this helps!
~kaikers

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3 years ago

When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed?

Ad libitum [116K]

You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.

For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.

6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2

For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.

13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2

As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.

In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!

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