Which of the following explains why a longer bond is also a weaker bond? Help plz

Answer:

Dipole-dipole interactions

Explanation:

Answer:

2)

[tex]\% Mg=20.2\%\\\\\% S=26.6\%\\\\\% O=53.2\%[/tex]

3)

[tex]\% Ag=93.1\%\\\\\% O=6.9\%[/tex]

Explanation:

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2) In this case, since magnesium sulfate is MgSO₄, we can see how magnesium weights 24.305 g/mol, sulfur 32.06 g/mol and oxygen 64.00 g/mol as there is one atom of magnesium as well as sulfur but four oxygen atoms for a total of g/mol; thus the percent compositions are:

[tex]\% Mg=\frac{24.305}{120.36 } *100\%=20.2\%\\\\\% S=\frac{32.06}{120.36 } *100\%=26.6\%\\\\\% O=\frac{64.00}{120.36 } *100\%=53.2\%[/tex]

3) In this case, although the element seems to contain Ag and O, we infer its molecular formula is Ag₂O; thus, since we have two silver atoms weighing 215.74 g/mol and one oxygen atom weighing 16.00 g/mol for a total of 231.74 g/mol, we obtain the following percent compositions:

[tex]\% Ag=\frac{215.74}{231.74} *100\%=93.1\%\\\\\% O=\frac{16.00}{231.74} *100\%=6.9\%[/tex]

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Answer:

c. fluorine

Explanation:

A single replacement reaction is defined as the chemical reaction in which a strong molecule replaces the weak molecule from a compound.

In a single replacement reaction, the chloride ion in NaCl can be replaced     by fluorine and gives Sodium fluoride as fluorine (F) is stronger than chlorine (Cl) and from other given molecules also. So the single replacement reaction between NaCl and F will be:

NaCl + F2 => NaF + Cl2

Hence, the correct answer is "c. fluorine".

Answer:

C) the number of moles of one reactant reacts completely with the moles of the other reactant.

Explanation:

it's right put it

Answer:

C

Explanation:

Answer:

Gravitational potential energy depends on basically two factor;

Explanation:

The distances between both the bodies, as well as the mass within each object, are factors that influence gravitational potential energy.

Answer:

the second answer choice i think

Explanation:

Since there is excess hydrogen, that means that the concentration of hydrogen is not limiting.

According to the balanced chemical equation:

every mole of N2 and every 3 moles of H2 we obtain 2 moles of NH3. Since we have the amount of N2 and excess of hydrogen, we only need to work with the amount of N2.

1 mol of N2 will produce 2 moles of NH3

so now we can get a conversation tool :

moles of N2 * (2 moles of NH3/ 1 mole of N2) = moles of NH3

(2.22 mol of N2 x 2) ÷ 1 = 4.44 mol of NH3

Explanation:

as the Hydrogen (acid) ion decreases there is a corresponding increase in the Hydroxide (basic) ion.

5.972 × 10^24 kg

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Answer:

5.972 × 10^24 kg

hope this helps

Answer:

D no. is the answer of your question

Answer:

4.584*10^23 molecules

Explanation:

Find molar mass of nitrogen and oxygen on periodic table.

find molar mass of nitrogen dioxide

use this to find moles of nitrogen dioxide

multiply by avagadros number

Answer:

The answer is A. Surface currents are caused by wind deep water currents are caused by difference in water density.

Answer:

D . Na

Explanation:

Answer:

531 mL is the volume in mL

Answer:

Hi there!

I hope you are having an outstanding day.

I believe the answer to your question is 531!

Please adress me if I am incorrect. :)

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     Pandora

Explanation:

Answer:

Mole fraction of KCl = 0.4056

Explanation:

We'll begin by calculating the number of mole of each compound. This can be obtained as follow:

For NaCl:

Mass NaCl = 0.564 g

Molar mass of NaCl = 58.44 g/mol

Mole of NaCl =?

Mole = mass /Molar mass

Mole of NaCl = 0.564 / 58.44

Mole of NaCl = 0.0097 mole

For KCl:

Mass KCl = 1.52 g

Molar mass of KCl = 74.55 g/mol

Mole of KCl =?

Mole = mass /Molar mass

Mole of KCl = 1.52 / 74.55

Mole of KCl = 0.0204 mole

For LiCl:

Mass LiCl = 0.857 g

Molar mass of LiCl = 42.39 g/mol

Mole of LiCl =?

Mole = mass /Molar mass

Mole of LiCl = 0.857 / 42.39

Mole of LiCl = 0.0202 mole

Next, we shall determine the total mole in the mixture. This can be obtained as follow:

Mole of NaCl = 0.0097 mole

Mole of KCl = 0.0204 mole

Mole of LiCl = 0.0202 mole

Total mole =?

Total mole = Mole of NaCl + Mole of KCl + Mole of LiCl

Total mole = 0.0097 + 0.0204 + 0.0202

Total mole = 0.0503 mole

Finally, we shall determine the mole fraction of KCl in the mixture. This can be obtained as follow:

Mole of KCl = 0.0204 mole

Total mole = 0.0503 mole

Mole fraction of KCl =?

Mole fraction of KCl = Mole of KCl /Total mole

Mole fraction of KCl = 0.0204 / 0.0503

Mole fraction of KCl = 0.4056

Answer:

.4 is correct

Explanation:

Answer:

0.0103 mol

Explanation:

Step 1: Given data

Step 2: Convert "m" to grams

We will use the conversion factor 1 g = 1000 mg.

200. mg × 1 g/1000 mg = 0.200 g

Step 3: Calculate the moles (n) of naproxen sodium

We will use the following expression.

n = m/M = 0.200 g / (252 g/mol) = 7.94 × 10⁻⁴ mol

Step 4: Calculate the moles of hydrogen in 7.94 × 10⁻⁴ moles of naproxen sodium

According to the chemical formula, the molar ratio of naproxen sodium to H is 1:13. The moles of H are 13/1 × 7.94 × 10⁻⁴ mol = 0.0103 mol.

Answer:

Nitrobenzene is too deactivated (by the nitro group) to undergo a Friedel-Crafts alkylation.

Explanation:

The benzene ring in itself does not easily undergo electrophilic substitution reaction. Some groups activate or deactivate the benzene ring towards electrophilic substitution reactions.

-NO2 ia a highly deactivating substituent therefore, Friedel-Crafts alkylation of nitrobenzene does not take place under any conditions.

This reaction scheme is therefore flawed because Nitrobenzene is too deactivated (by the nitro group) to undergo a Friedel-Crafts alkylation.

Answer:

0.500-Molarity solution

Explanation:

The moles of the compound is given as the number of atomic mass unit in the compound. The moles of potassium nitrate in the solution are 0.6 mol

The molarity is the concentration unit, and it can be defined as the moles of compound present in the liter of solution.

The molarity can be expressed as:

[tex]\rm Molarity=\dfrac{Moles}{Volume(L)}[/tex]

The given potassium nitrate solution has, molarity = 2 M

The volume of the solution is 0.3 L.

Substituting the values for the moles of the compound:

[tex]\rm 2\;M=\dfrac{Moles}{0.3\;L} \\\\Moles=2\;\times\;0.3\;mol\\Moles=0.6\;mol[/tex]

The moles of potassium nitrate in 2 M solution is 0.6 mol.

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Answer:

Chalk is does not dissolve in water

Explanation:

i had the same question as you

Answer:

Explanation: There is no molecule or ion as ClO3. The correct formula of chlorate ion is ClO3-. bolivianouft and 4 more users found this answer helpful.

Explanation:

Answer:C

Explanation:

Answer:

Deer, Eagle, Raccoon, frog, turtle, fish, dragonfly.

Explanation:

They interact from the life cycle they eat each other...

Answer:

Joan's hypothesis should make a prediction about the answer to the question. A hypothesis is just an educated guess on what you think the outcome will be on the experiment. The prediction must be testable and stated in if-then form. For example, a good hypothesis for Joan's experiment could be, I think that 1/2 cup of salt will make the water boil quicker, than water without salt.

Explanation:

Hope it helped!

Answer:

A^4 +b^4

(a²)²+(b²)²

(a²+b²)²-2a²b²

or

(a²-b²)²+2a²b²

Answer:

Explanation:

Molarity of NaOAc needed

Using the Henderson-Hasselbalch Equation calculate base molarity needed given [HOAc] = 1.00M and pKa(NaOAc) = 4.75 and [HOAc] = 1.00m.

pH = pKa + log [NaOAc]/[HOAc]

5.00 = 4.75 + log[NaOAc]/[1.00M]

5.00 - 4.75 = log [NaOAc] - log[1.00M]

log [NaOAc] = 0.25 => [NaOAc] = 10⁰·²⁵ M = 1.78

Given 10ml of HOAc, how much (ml) 1.78M NaOAc to obtain a buffer pH of 5.00.

Determine Volume of Base Needed

(M·V)acid = (M·V)base => V(base) = (M·V)acid / (M)base

Vol (NaOAc) needed = (1.00M)(0.010L)/(1.78M) = 0.0056 liter = 5.6 ml.

Checking Results:

5.00 = 4.75 + log [1.78M]/[1.00M] = 4.75 + 0.25 = 5.00    QED.

The volume of 1.00 M sodium acetate solution needed to prepare an acetic/acetate buffer of pH 5.00 using 10.0 mL of 1.00M acetic acid solution is 17.8 mL.

We can find the volume of the acetate solution with the Henderson-Hasselbalch equation:

[tex]pH = pka + log(\frac{[CH_{3}COONa]}{[CH_{3}COOH]})[/tex]   (1)  

Where:

[CH₃COOH] = 1.00 M                      

[CH₃COONa] =?

pH = 5.00

pKa = 4.75

From equation (1), we have:

[tex] log(\frac{[CH_{3}COONa]}{[CH_{3}COOH]}) = pH - pKa [/tex]

[tex] \frac{[CH_{3}COONa]}{[CH_{3}COOH]} = 10^{pH - pKa} [/tex]

[tex] \frac{[CH_{3}COONa]}{[CH_{3}COOH]} = 10^{5.00 - 4.75} = 1.78 [/tex]

Now, the volume of the acetate solution is:

[tex]\frac{n_{CH_{3}COONa}/Vt}{n_{CH_{3}COOH}/Vt} = 1.78[/tex]  

Since the total volume is the same, we have:

[tex]\frac{n_{CH_{3}COONa}}{n_{CH_{3}COOH}} = 1.78[/tex]  

[tex] \frac{[CH_{3}COONa]_{i}*V_{b}}{[CH_{3}COOH]_{i}*Va} = 1.78 [/tex]  

Solving for Vb

[tex] Vb = \frac{1.78*[CH_{3}COOH]_{i}*Va}{[CH_{3}COONa]_{i}} = \frac{1.78*1.00M*10.0mL}{1.00 M} = 17.8 mL [/tex]

Therefore, we need to add 17.8 mL of sodium acetate solution.

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