5g of a mixture of KOH and KCl with water form a solution of 250mL. We have 25ml of this solution and we mix it with 14,3mL of HCl 0,1M. What's the % of the inicial mixture?The 14,3mL of HCl are necessary to neutralize the KOH.

What are the reactions and what do I need to calculate?

Answers

Answer 1

Answer: we know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl

Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.

To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH

Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572. That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH

Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH

Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH

I hope this helps.

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What hydrocarbons burn completely in an excess of oxygen, the products are A) carbon monoxide and water B) carbon dioxide and water C) carbon monoxide and carbon dioxide D) carbon dioxide and carbon

Answers

The right answer for the question that is being asked and shown above is that: "C) carbon monoxide and carbon dioxide" hydrocarbons burn completely in an excess of oxygen, the products are C) carbon monoxide and carbon dioxide

Answer:

C just took the test

Explanation:

How do valence electrons affect reactivity?

Answers

The less valence electrons there are, the less stable the atom becomes, causing it to react with other atoms. More valence electrons adds mass to the atom, making it less reactive

At STP, which 2.0-gram sample of matter uniformly fills a 340-milliliter closed container?

Answers

Answer;

D, Xe (g)

Solution and explanation;

If 2g has a volume of 340ml.

Density is 1000/340*2 = 5.88g/litre.

-This rules out the two solids, choices 2) &3)

If 1 litre has mass 5.88g,

then 22.4 liters (volume at STP) has mass 5.88*22.4 = 131.8g/mol

molar mass Br2 = 80*2 = 160g/mol NO

molar mass Xe = 131.3g/mol = YES.

Answer is Xe

Final answer:

To determine the matter filling a closed container at STP given a 2.0 gram sample, you can use the ideal gas law equation to find the number of moles. Then, divide the number of moles by the volume and the gas constant to solve for the pressure. Compare the pressure obtained to known substances' vapor pressures at STP to identify the matter.

Explanation:

The question is asking for the specific type of matter that would uniformly fill a 340-milliliter, closed container at STP (Standard Temperature and Pressure) when given a 2.0-gram sample. To determine the matter, we can use the ideal gas law equation, PV = nRT, and rearrange it to solve for n, the number of moles. Then, we can use the molar mass of the substance to find its identity.

First, convert the volume from milliliters to liters by dividing it by 1000: 340 mL ÷ 1000 = 0.34 L. Next, convert the mass from grams to moles using the molar mass of the substance:

1.(Conversion factor) Given: 2.0 g sample, 1 mole = molar mass

2.(Calculation) Moles of substance = 2.0 g ÷ molar mass

Once you have the number of moles, divide it by the volume (in liters) and the universal gas constant (0.0821 L·atm/mol·K) and solve for the pressure:

1.(Ideal Gas Law) PV = nRT

2.(Substitution) P × 0.34 L = n × 0.0821 L·atm/mol·K

3.(Isolation) P = (n × 0.0821 L·atm/mol·K) ÷ 0.34 L

After solving for the pressure, compare it to known substances' vapor pressures at STP to determine the identity of the matter in the container.

Learn more about Ideal Gas Law here:

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The smallest unit into which a compound can be divided and still be that same compound is a(n) ___.

Answers

The molecule of that compound! Hope this helps!

Answer:

The molecule of that compound.

Explanation:

17 Two substances in a mixture differ in density and particle size. These properties can be used to(1) separate the substances(2) chemically combine the substances
(3) determine the freezing point of the mixture
(4) predict the electrical conductivity of the mixture

Answers

Answer:The correct answer is option (1).

Explanation:

Density is defined as mass of substance present in the unit volume of the substance.

Density can be used to separate two substance from their mixture. Substance with lower density will place itself on the top of the substance with higher density, And in order to distinguish those two substance from each other one can easily observe the boundary among those substance.

If they're different sizes and densities, you are able to separate the substances.

How does math relate to electrons and ion formation?

Answers

In order to find the NET nuclear charge from an atom's valence electron to the proton nucleus, you need to do some simple math in order to find the charge.

You would also need to apply math when you need to figure out what the charge on an ion is by either adding or taking away electrons depending on whether it is an anion or cation.

Answer and explanation;

Ions are formed when an atom looses or gains electrons.

-Ionic charges are created when there is an imbalance of protons and electrons on an atom. Since we can not change the number of protons on the nucleus of an atom, a positively charged ion will be created when there are fewer electrons than protons (in other words when an atom looses electron). A negatively charged ion is formed or created when there are more electrons than protons in an atom (when an atom gains electrons).

-When electron are added, you are adding more negative charge and thus, results in the creation of a negative ion (anion), and when electrons are being lost by an atom, one is removing negative charge, resulting in creation of a positive ion (cation).

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