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g100num [7]
3 years ago
4

Classify each substance based on the intermolecular forces present in that substance.A. H2OB. CH4C. CO D. CH3Cl1. Hydrogen bondi

ng, dipole dipole, and dispersion2. Dipole dipole and dispersion only 3. Dispersion only
Chemistry
1 answer:
Fynjy0 [20]3 years ago
5 0

Answer:

H2O - Hydrogen bonding, dipole dipole, and dispersion

CH4 - Dispersion only

CO - Dipole dipole and dispersion only

CH3Cl - Dipole dipole and dispersion only

Explanation:

Hydrogen bonding can only exist when hydrogen is bonded to a small highly electronegative atom such as oxygen hence hydrogen bonds, dipole interactions and dispersion by are present in water.

CH4 is a nonpolar molecule hence only dispersion forces are present.

CH3Cl and CO both possess dipoles in the molecule hence both dipole interactions and dispersion forces exist in the molecule.

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PLEASE HELP i’ll mark brainliest ( reporting fake answers)
kherson [118]

Delta S reaction= Delta S products- Delta S reactants

don't forget to mulitiply by coefficients

also

here is a really slow way to do it

you know the moles of gas increased

so Delta S is positive

so its B or D

then just do the units digit to see which one match up

6 0
3 years ago
How many protons and neutrons does nitrogen-14 have?
zhenek [66]

Answer:

7 protons

Isotopes are atoms that have the same number of protons but different numbers of neutrons in the nucleus. You know that nitrogen-14 has 7 protons in the nucleus because it is an isotope of nitrogen, which has an atomic number equal to 7 .

4 0
4 years ago
1. According to the equation, what mass of hydrogen fluoride is necessary to produce 2.3 g of sodium fluoride?
Luba_88 [7]

Answer:

1.09 grams

Explanation:

According to the following chemical equation:

HF + NaNO₃ -> HNO₃ + NaF

1 mol of hydrogen fluoride (HF) produces 1 mol of sodium fluoride (NaF). Thus, we first convert from mol to grams by using the molar mass (MM) of each compound:

MM(HF)= (1 g/mol x 1 H) + (19 g/mol x 1 F) = 20 g/mol HF

1 mol HF x 19.9 g/mol HF = 20 g

MM(NaF) = (23 g/mol x 1 Na) + (19 g/mol x 1 F) = 42 g/mol NaF

1 mol NaF x 42 g/mol NaF = 42 g

Thus, from 20 g of HF are produced 42 g of NaF  ⇒ 20 g HF/42 g NaF. We multiply this stoichiometric ratio by the mass of NaF produced to calculate the required mass of HF:

20 g HF/42 g NaF x 2.3 g NaF = 1.09 g HF

Therefore, 1.09 grams of HF are necessary to produce 2.3 g of NaF.

5 0
3 years ago
1. convert 8.43 x 1025 atoms of Fe to moles
Luba_88 [7]

Answer:

The number of moles of Fe are, 139.9 moles.

The number of moles of CO are, 7.89 moles.

The number of moles of Al are, 4.04\times 10^{-4} moles.

Explanation :

As we know that 1 mole of substance contains 6.022\times 10^{23} atoms or molecules.

Part 1:

As, 6.022\times 10^{23} atoms of Fe present in 1 mole

So, 8.43\times 10^{25} atoms of Fe present in \frac{8.43\times 10^{25}}{6.022\times 10^{23}}\times 1=139.9 mole

Thus, the number of moles of Fe are, 139.9 moles.

Part 2:

As, 6.022\times 10^{23} molecules of CO present in 1 mole

So, 4.75\times 10^{24} molecules of CO present in \frac{4.75\times 10^{24}}{6.022\times 10^{23}}\times 1=7.89 mole

Thus, the number of moles of CO are, 7.89 moles.

Part 3:

As, 6.022\times 10^{23} atoms of Al present in 1 mole

So, 2.43\times 10^{20} atoms of Al present in \frac{2.43\times 10^{20}}{6.022\times 10^{23}}\times 1=4.04\times 10^{-4} mole

Thus, the number of moles of Al are, 4.04\times 10^{-4} moles.

4 0
3 years ago
During an experiment, 104 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of
postnew [5]

Answer:

92.4 grams.

Explanation:

  • From the balanced reaction:

<em>CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O,</em>

1.0 mole of CaCO₃ reacts with 2.0 moles of HCl to produce 1.0 mole of CaCl₂, 1.0 mole of CO₂, and 1.0 mole of H₂O.

  • We need to calculate the no. of moles of (104 g) of CaCO₃:

<em>no. of moles of CaCO₃ = mass/molar mass</em> = (104 g)/(100.08 g/mol) = <em>1.039 mol.</em>

<u><em>Using cross multiplication:</em></u>

1.0 mole of CaCO₃ produce → 1.0 mole of CaCl₂.

∴ 1.039 mole of CaCO₃ produce → 1.039 mole of CaCl₂.

∴ The amount of CaCl₂ produced = no. of moles x molar mass = (1.039 mol)(110.98 g/mol) = 114.3 g.

∵ percent yield of the reaction = [(actual yield)/(theoretical yield)] x 100.

Percent yield of the reaction = 80.15%, theoretical yield = 115.3 g.

<em>∴ actual yield = [(percent yield of the reaction)(theoretical yield)]/100 </em>= [(80.15%)/(115.3 g)] / 100 = <em>92.42 g ≅ 92.4 g.</em>

7 0
3 years ago
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