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3 years ago

The properties of each state of a substance are due largely to the degree (amount) of intermolecular bonding.

Chemistry

2 answers:

sertanlavr [38]3 years ago

8 0

Answer: true (not false)

Viktor [21]3 years ago

5 0

Answer:

true

Explanation:

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What is the electron configuration for S-

Sonbull [250]

Answer:

The Electron Configuration for Sulfur(S)= 1s^2 2s^2 2p^6 3s^2 3p^4

Explanation:

Use the sheet attached to help with a future question. (Hint: Follow the red lines). The atomic number will be equal to the number of powers raised.

4 0

4 years ago

You determine that Lauren's EER is 2200 kcal/day. You recommend and intake of 60% CHO, 15% protein, and 25% fat. Approximately h

const2013 [10]

Answer:

The answer to your question is 330 g of CHO

Explanation:

Data

Calories needed = 2200 kcal/day

CHO = 60%

Proteins = 15%

Fats = 25%

Grams of carbohydrates needed = ?

Process

1.- Calculate the number of calories in 60% of 2200 kcal

2200 kcal ---------------- 100%

x --------------- 60%

x = (60 x 2200) / 100

x = 1320 kcal

2.- Calculate the grams of CHO

1 g of CHO ---------------- 4 kcal

x ---------------- 1320 kcal

x = (1320 x 1) / 4

x = 1320/4

x = 330 g of CHO

6 0

3 years ago

In the balanced equation, 4KO2 + 2CO2 = 2K2CO3 + 3O2 which two chemicals have a molar ratio of 3:4?

Sveta_85 [38]

4KO₂ + 2CO₂ → 2K₂CO₃ + 3 O₂

The two chemicals that have a molar ratio of 3:4 is in equation above O₂ and KO₂

<em><u>explanation</u></em>

The molar ratio is the ratio between the amount in moles of two compounds that are involved in a chemical reaction.

In the equation above the mole ratio is determined by examining coefficient ( <em>number in front of the formula</em>)

the coefficient of O₂ is 3 while that of KO₂ is 4 therefore the mole ratio of O₂:KO₂ is 3:4

3 0

3 years ago

Read 2 more answers

a sample of natural gas contains 8.24 moles of methane, 0.421 moles of ethane, and 0.116 moles of propane. if the total pressure

Varvara68 [4.7K]

The partial pressure (Px) of a gas in a gas mixture is equal to its mole fraction (Xi) multiplied by the total pressure (P) of the gas mixture. That means that we have to calculate the mole fraction of each gas, then calculate its partial pressure. The mole fraction of a gas is its number of moles (n) divided by the total number of moles.

$$ Mole fraction of methane: \\$\chi_{\text {methane }}=\frac{\mathrm{n}_{\text {methane }}}{\mathrm{n}_{\text {total }}} \chi_{\text {methane }}=\frac{8.24 \mathrm{~mol}}{8.24 \mathrm{~mol}+0.421 \mathrm{~mol}+0.116 \mathrm{~mol}} \chi_{\text {methane }}=\frac{8.24 \mathrm{~mol}}{8.78 \mathrm{~mol}} \chi_{\text {methane }}=0.938$

$$Partial Pressure of methane:\\$\mathrm{P}_{\text {methane }}=\chi_{\text {methane }} \times \mathrm{PP}_{\text {methane }}=0.938 \times 1.37 \mathrm{~atm} \mathbf{P}_{\text {methane }}=\mathbf{1 . 2 8} \mathbf{~ a t m}$

$$Mole fraction of ethane: \\$\chi_{\text {ethane }}=\frac{\mathrm{n}_{\text {ethane }}}{\mathrm{n}_{\text {total }}} \chi_{\text {ethane }}=\frac{0.421 \mathrm{~mol}}{8.78 \mathrm{~mol}} \chi_{\text {ethane }}=0.0479$

$$Partial pressure of ethane:\\$\mathrm{P}_{\text {ethane }}=\chi_{\text {ethane }} \times \mathrm{PP}_{\text {ethane }}=0.0479 \times 1.37 \mathrm{~atm} \mathrm{P}_{\text {ethane }}=\mathbf{0 . 0 6 5 6} \mathbf{~ a t m}$

$$Mole fraction of propane:\\$\chi_{\text {propane }}=\frac{\mathrm{n}_{\text {propane }}}{\mathrm{n}_{\text {total }}} \chi_{\text {propane }}=\frac{0.116 \mathrm{~mol}}{8.78 \mathrm{~mol}} \chi_{\text {propane }}=0.0132$

<h3>What is Dalton’s Law?</h3>

Dalton's law of partial pressures is a gas law that states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures exerted by each individual gas in the mixture. The mole fraction of a given gas in a gas mixture is equal to the ratio of the partial pressure of that gas to the total pressure exerted by the gas mixture. This mole fraction can also be used to calculate the total number of moles of constituent gas if the total number of moles of the mixture is known. In addition, the mole fraction can also be used to calculate the volume of a certain gas in a mixtur.

To learn more about Dalton’s Law, visit:

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6 0

1 year ago

Consider the following data. NO2(g) equilibrium reaction arrow identifying the requirement for light NO(g) + O(g) K = 6.8 ✕ 10-4

LenKa [72]

Answer:

$K=3.9*10^{-82}$