Which of the following statements is true about one formula unit of RuF2?<br>

A sample of gas has an initial volume of 20 L and an initial pressure of 2.5 atm. If the pressure changes to 3.1 atm, what is th

Studentka2010 [4]

The new volume is mathematically given as

V2= 16.12 L

<h3>What is the new volume?</h3>

Question Parameters:

A sample of gas has an initial volume of 20 L

and an initial pressure of 2.5 atm.

If the pressure changes to 3.1 atm

Generally, the equation for Pressure is mathematically given as

P1V1 = P2 V2

2.5 * 20 = 3.1 * V2

V2 = 50/3.1

V2= 16.12 L

For more information on volume

brainly.com/question/1578538

6 0

1 year ago

The accepted value is 29.35. Which correctly describes this student's experimental data?

kkurt [141]

Answer:

- Both accurate and precise.

Explanation:

The experimental data is characterized as the data that is generated through a specific test or measurement with the purpose of analysis or evaluation. As per the question, the minuteness of the given data(29.27, 29.32, and 29. 27)most aptly describes that the experimental data collected by the student possesses 'accuracy and precision' apt and specific. It would yield the most worthy and reliable conclusion.

3 0

2 years ago

Why was UwU createdwwwwwwwwwwwwwwwwwwwwwwwwwwwww

photoshop1234 [79]

a typable representation of emotion

5 0

2 years ago

A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.692 kg of water

S_A_V [24]

Answer:

494.49 g of NaCl.

Explanation:

Data obtained from the question include the following:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mass of NaCl =.?

Next, we shall determine the number of mole of NaCl in the solution.

Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as

Molality = mole of solute /Kg of solvent

With the above formula, we can obtain the number of mole NaCl in the solution as follow:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mole of NaCl =..?

Molality = mole of solute /Kg of solvent

3.140 = mole of NaCl /2.692

Cross multiply

Mole of NaCl = 3.140 x 2.692

Mole of NaCl = 8.45288 moles

Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:

Mole of NaCl = 8.45288 moles

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =.?

Mole = mass /Molar mass

8.45288 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 8.45288 × 58.5

Mass of NaCl = 494.49 g.

Therefore, 494.49 g of NaCl are present in the solution.

8 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

2 years ago

Which of the following statements is true about one formula unit of RuF2?​

Which of the following statements is true about one formula unit of RuF2?