At room temperature, oxygen is a gas and hydrogen is a gas. When they combine at room temperature, they form a _______________.

Answer:

Dipole-dipole interactions

Explanation:

Answer:

D no. is the answer of your question

Answer:

Ethanol is another alcohol, and since there are less carbons in its chain, it will be more polar. So Ethanol is the most polar bond, it has two different types of bonding between its constituent atoms. The bonds between the hydrogen and carbon atoms are nonpolar covalent bonds. The hydrogen-oxygen and carbon-oxygen bonds are polar covalent bonds.

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:

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. :)

Much love,

     Pandora

Explanation:

Answer:

the second answer choice i think

Explanation:

Answer:C

Explanation:

Answer:

A^4 +b^4

(a²)²+(b²)²

(a²+b²)²-2a²b²

or

(a²-b²)²+2a²b²

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:

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:

I'm confused I think you're suppose to put a picture if you want someone to answer your question just saying.

Explanation:

Answer:

The understanding of the earth, air, fire, and water as the basic components of matter.

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:

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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I hope it helps you!  

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

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:

D . Na

Explanation:

Answer:

CH4

Explanation:

if it is wrong, i blame my brain

Answer:

Chalk is does not dissolve in water

Explanation:

i had the same question as you

Answer:

tysm!!!! :D <<<33333

Explanation:

5.972 × 10^24 kg

hope it helps

Answer:

5.972 × 10^24 kg

hope this helps

Answer:

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

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:

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:

The materials which allow electrons to flow through them are Semiconductors and conductors. Hence Option (B) is correct.

A conductor, or electrical conductor, is a substance or material that allows electricity to flow through it.

In a conductor, electrical charge carriers, usually electrons or ions, move easily from atom to atom when voltage is applied.

Therefore, The materials which allow electrons to flow through them are Semiconductors and conductors. Hence Option (B) is correct.

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Answer:semiconductors and conductors

Explanation: