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wolverine [178]
2 years ago
14

a researcher finds 17grams of bromine will react with 3.404g of oxygen respectively. what is the mass of oxygen per one gram of

bromine​
Chemistry
1 answer:
GenaCL600 [577]2 years ago
6 0

Answer:

\frac{m_{O}}{m_{Br}}=0.2

Explanation:

Hello there!

In this case, according to the description, it turns out possible for us to calculate the required ratio by performing the following ratio:

\frac{m_{O}}{m_{Br}}

Thus, we plug in the given masses to obtain:

\frac{m_{O}}{m_{Br}}=\frac{3.404g}{17g}\\\\\frac{m_{O}}{m_{Br}}=0.2

Regards!

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Methane burns in oxygen to produce carbon dioxide and water. Which of the following represents
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A

Explanation:

CH4+O2-CO2+ H20

that mean methane has burn in oxygen to produce CO2

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What information is needed to determine the energy of an electron in a many-electron atom? a) n. b) l. c) ml. d)ms
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Answer:

The correct answer is b.

Explanation:

The quantum number n specifies the energetic level of the orbital, the first level being the one with the least energy. As n increases, the probability of finding the electron near the nucleus decreases and the orbital energy increases.

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a sample of 3.00 g of so2 (g)originally in a 5.00 L vesselat 21 degee Celsius is transferred to a 10.0 L vessel at 26 degree Cel
eimsori [14]

Answer:

1) The partial pressure of SO₂ gas in the larger container = 0.115 atm.

2) The partial pressure of N₂ gas in the larger container = 0.206 atm.

3) The total pressure in the vessel = 0.321 atm.

Explanation:

  • To calculate the partial pressure of each gas, we can use the general law of ideal gas: PV = nRT.

where, P is the partial pressure of the gas in atm,

V is the volume of the vessel in L,

n is the no. of moles of the gas,

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K.

<u><em>1) What is the partial pressure of SO₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (3.0 g)/(64.066 g/mol) = 0.047 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.047 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.115 atm.

<u><em>2) What is the partial pressure of N₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (2.35 g)/(28.0 g/mol) = 0.084 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.084 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.206 atm.

<u><em>3) What is the total pressure in the vessel?</em></u>

  • According to Dalton's law the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

<em>∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.</em>

∴ The total pressure in the vessel = 0.115 + 0.206 = 0.321 atm.

5 0
3 years ago
During laparoscopic surgery , carbon dioxide gas is used to expand the abdomen to help create a larger working space. If 4.80 L
Studentka2010 [4]

Answer:

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Explanation:

To solve this problem we need to use the PV=nRT equation.

First we <u>calculate the amount of CO₂</u>, using the initial given conditions for P, V and T:

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1.03 atm * 4.80 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 291.16 K

We <u>solve for n</u>:

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Then we use that value of n for another PV=nRT equation, where T=37 °C (310.16K) and P = 745 mmHg (0.98 atm).

  • 0.98 atm * V = 0.207 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 310.16 K

And we <u>solve for V</u>:

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