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Enter a chemical equation for NaOH(aq) showing how it is an acid or a base according to the Arrhenius definition. Consider that

yuradex [85]

Answer:

NaOH(aq) is a Base.

Explanation:

Those substances which give or release $OH^{-}$ ions in aqueous solution are called as the Arrhenius Bases.

In the aqueous solution, NaOH dissociates as follows -

$NaOH^{}$ ↔ $Na^{+} + OH^{-}$

If it reacts with a strong acid HCl, the chemical equation for this reaction will be as follows -

$HCl + NaOH = NaCl + H_{2} O$

6 0

3 years ago

How many molecules of sulfur dioxide are present in 1.60 moles of sulfur dioxide

ELEN [110]

For this question, you must know that there are 6.022e23 atoms/molecules per mole of any substance (this is Avogadro's number). Therefore, your answer is 6.022e23 * 1.60 = 9.64e23 molecules of sulfur dioxide. (the "e" represents "times ten to the power of ___ ")

5 0

3 years ago

A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

Ede4ka [16]

<u>Answer:</u> The mass of glucose in final solution is 1.085 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL$

Putting values in above equation, we get:

$0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g$

Hence, the mass of glucose in final solution is 1.085 grams

4 0

3 years ago

The ionization constant (Ka) of HF is 6.7 X 10 ^-4. Which of the following is true in a 0.1 M solution of this acid?

spin [16.1K]

The ionization equation is:

HF ⇄ H(+) + F(-)

The ionization constant is Ka = [H(+)] * [H(-)] / [HF]

=> [H(+)] * [F(-)] = Ka * [HF]

Given that Ka < 1

[H(+)] * [F(-)] < [HF]

Which is [HF] > [H(+)] * [F(-)] the option a. fo the list of choices.

8 0

3 years ago

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Which product is used on the natural nail prior to product application to assist in adhesion and serves to chemically bond the e

ira [324]

The product that is used on the natural nail prior to application to assist in adhesion and serves to chemically bond the enhancement product to the natural nail is known as nail primer.

<h3>What is a nail primer?</h3>

A nail primer is a chemical agent used in esthetic centers before applying a colored polish to the nails and serves as an adhesive product.

The nail primers are also very useful for improving the cleaning efficiency of the product before its application.

Nail care products include different types of chemical formulations such as, for example, creams that reinvigorate the cuticle.

In conclusion, the chemical formulation employed on the natural nail that is capable of enhancing and also assisting adhesion is called the nail primer.

Learn more about nail esthetic products here:

brainly.com/question/14498053

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8 0

2 years ago