Ask question

Ask question

All categories

4 years ago

9

Chemical name for S2F3

1 answer:

Margarita [4]4 years ago

3 0

Answer:

Sulfur tetrafluoride

Explanation:

You might be interested in

A sample of gas has its number of molecules quadrupled, its Kelvin temperature doubled, and its volume tripled. By what factor h

steposvetlana [31]

The new pressure changed relative to the original by a factor of 8/3.

<h3>What is the change in the pressure?</h3>

Using the ideal gas equation;

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

P1 = nRT/V

P2 = 4n * R * 2T/3V

Hence;

P2/P1 = 4n * R * 2T/3V ÷ nRT/V

P2/P1 = 4n * R * 2T/3V * V/nRT

P2/P1 =4 * 2/3

= 8/3

Learn more about ideal gas law:brainly.com/question/13821925

#SPJ1

4 0

2 years ago

Calculate the mass of 6.9 moles of nitrous acid (HNO2). Explain the process or show your work by including all values used to de

AlexFokin [52]

Answer:- 324.3 grams.

Solution:- We have been given with 6.0 moles of nitrous acid and asked to calculate it's grams. Moles to grams is a unit conversion and for doing this conversion we multiply the given moles by the molar mass of the compound.

Molar mass is the formula mass and to calculate this the atomic masses of each atom are multiplied by their respective subscripts that is the number of the atom in the compound .

For example, $NO_2$ has one nitrogen atom and one oxygen atom. So, the molar mass of this is = atomic mass of N + 2(atomic mass of O)

= 14 + 2(16)

= 14 + 32

= 46 gram per mol

gram per mol is the unit of molar mass. So, the molar mass of $NO_2$ is 46 grams per mol.

Let's calculate the molar mass of nitrous acid using the same concept.

molar mass of [ $HNO_2$ = 1 + 14 + 2(16)

= 1 + 14 + 32

= 47 grams per mol

Now, 6.9 moles of nitrous acid could easily be converted to grams as:

$6.9molHNO_2(\frac{47g}{1mol})$

= 324.3 g

Hence, the mass of 6.9 moles of nitrous acid is 324.3 grams.

6 0

3 years ago

Gasoline is a mixture of compounds. A gasoline sample was found to be 2.851% (by mass) octane (C₈H₁₈). 482.6 g of gasoline is co

Pani-rosa [81]

Answer:

See below.

Explanation:

The mass of octane in the sample of gasoline is 0.02851 * 482.6 = 13.759 g of octane.

The balanced equation is:

2C8H18(l) + 25O2(g) ----> 16CO2(g) + 18H2O(g)

From the equation, using atomic masses:

228.29 g of octane forms 704 g of CO2 and 324.3 g of H2O

So the mass of CO2 formed from the combustion of 13.759 g of octane = (704 * 13.759) / 228.29

= 42.43 g of CO2.

Amount of water = 324.3 * 13.759) / 228.29

= 19.55 g of H2O.

5 0

3 years ago

The graph below shows how the temperature and volume of a gas vary when

Sever21 [200]

Answer:

C

Explanation:

because it remains the same

3 0

3 years ago

You are using a standardized 0.1250 M solution of sodium hydroxide to determine the concentration of sulfuric acid. After pipett

Sergeu [11.5K]

Answer: Thus the concentration of the sulfuric acid is 0.1201 M

Explanation:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?\\V_1=10.00mL\\n_2=1\\M_2=0.1250M\\V_2=19.22mL$

Putting values in above equation, we get:

$2\times M_1\times 10.00=1\times 0.1250\times 19.22\\\\M_1=0.1201M$

Thus the concentration of the sulfuric acid is 0.1201

4 0

3 years ago