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3 years ago

Which mass of gas would occupy a volume of 3dm3 at 25°C and 1 atmosphere pressure?

Chemistry

1 answer:

balu736 [363]3 years ago

8 0

Answer:

C 8.09 SO2 gas

Explanation:

As we have the volume (3dm³ = 3L), temperature (25°C + 273 = 298K), and pressure (1atm), we can solve to moles of gas using:

PV = nRT

PV / RT = n

1atm*3L / 0.082atmL/molK*298K =¨

0.123 moles of gas you have.

Now, to convert these moles to mass we use molar mass (32g/mol for O2, 28g/mol for N2, 64g/mol for SO2, and 44g/mol for CO2).

Mass of 0.123 moles of these gases is:

O2 = 0.123 moles * 32g/mol = 3.94g of O2. A is wrong

N2 = 0.123 moles * 28g/mol = 3.4g of N2. B is wrong

SO2 = 0.123 moles * 64.1g/mol = 7.9g of SO2≈ 8.09g of SO2, C is possible

CO2 = 0.123 moles * 44g/mol = 5.4g of CO2. D is wrong

Right answer is:

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At sea level water blank at 100 degrees celceius

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Boils

Celsius is based on the freezing and boiling points of water.

Water freezes below 0 degrees Celsius (32 degrees Fahrenheit), and it boils above 100 degrees Celsius (212 degrees Fahrenheit).

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2 years ago

What is 6421 in scientific notation

Lynna [10]

The scientific notation should specific the significant digits. So it to write like x.xx * 10 ^ x. So the answer is 6.421 * 10 ^3.

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3 years ago

What is the molarity of a HCl solution, if 28.6 mL of a 0.175 m NaOH solution is needed to neutralize a 25.0 mL sample of the HC

Katen [24]

The molarity of the HCl solution needed to neutralize 28.6 mL of a 0.175 M NaOH solution is 0.2002 M

We'll begin by writing the balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H₂O

From the balanced equation above,

The mole ratio of the acid, HCl (nA) = 1

The mole ratio of the base, NaOH (nB) = 1

From the question given above, the following data were obtained:

Volume of base, NaOH (Vb) = 28.6 mL

Molarity of base, NaOH (Mb) = 0.175 M

Volume of acid, HCl (Va) = 25 mL

<h3>Molarity of acid, HCl (Ma) = ?</h3>

The molarity of the acid, HCl can be obtained as follow:

MaVa / MbVb = nA / nB

(Ma × 25) / (0.175 × 28.6) = 1

(Ma × 25) / 5.005 = 1

Cross multiply

Ma × 25 = 5.005

Divide both side by 25

Ma = 5.005 / 25

Therefore, the molarity of the acid, HCl needed for the reaction is 0.2002 M

Learn more: brainly.com/question/25573711

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3 years ago

What is an ion and how is it formed?

grin007 [14]

Answer: Ions are an atom or molecule with a net electric charge due to the loss or gain of one or more electrons. They are formed when atoms lose or gain electrons in order to fulfill the octet rule and have full outer valence electron shells.

Explanation:

When they lose electrons, they become positively charged and are named cations. When they gain electrons, they are negatively charged and are named anions.

6 0

3 years ago

Considering the volumes and the concentrations used in mixture 1, what percentage of the moles of H2O2 present have been consume

Kisachek [45]

Answer:

1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.

Explanation:

the total volume of the mixture is equal to:

Vmix = 75 + 30 + 25 + 5 + 5 + 10 = 150 mL

the moles of each species in the mix equals:

Na2S2O3 = 0.05 * 0.005 = 0.00025 moles

KI = 0.05 * 0.025 = 0.00125 moles

H2O2 = 1.02 * 0.01 = 0.0102 moles

the following equation shows the reaction between I2 and S2O32:

I2 + S2O32 = 2I- + S4O62-

The same way:

2I- + 2H+ + H2O2 = I2 + 2H2O

1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

3 0

3 years ago