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Dafna11 [192]
3 years ago
5

Holly has an unknown substance in a beaker. she wants to determine the relative ph of the unknown substance. she places a piece

of blue litmus paper into the substance, and the litmus paper turns red. the substance in the beaker
Chemistry
2 answers:
FrozenT [24]3 years ago
8 0

Answer: its c

Explanation:

i have study island to :/

AURORKA [14]3 years ago
6 0

Blue litmus paper turns red in the presence of an acid. Therefore, it can be assumed that the substance in the beaker is an acid.

Acids have a pH level of less than 7. Consequently, it can be assumed that the substance has a pH level less than 7.

You might be interested in
(2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) ) If 165 mL of oxygen is produced at 30.0 °C and 90.0 kPa, what mass of KClO3 was decomposed
soldier1979 [14.2K]

Taking into account the reaction stoichiometry and ideal gas law, 0.48144 grams of KClO₃ was decomposed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 KClO₃  → 2 KCl + 3 O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • KClO₃: 2 moles  
  • KCl: 2 moles
  • O₂: 3 moles

The molar mass of the compounds is:

  • KClO₃: 122.45 g/mole
  • KCl: 74.45 g/mole
  • O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • KClO₃: 2 moles ×122.45 g/mole= 244.8 grams
  • KCl: 2 moles ×74.45 g/mole= 148.9 grams
  • O₂: 3 moles ×32 g/mole= 96 grams

<h3>Ideal gas law</h3>

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

  • P is the gas pressure.
  • V is the volume that the gas occupies.
  • T is the temperature of the gas.
  • R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
  • n is the number of moles of the gas.

<h3>Number of O₂ produced.</h3>

165 mL of oxygen is produced at 30.0 °C and 90.0 kPa. This is, you know:

  • P= 90 kPa= 0.888231 atm (being 101.325 kPa= 1 atm)
  • V= 165 mL= 0.165 L (being 1000 mL= 1 L)
  • n= ?
  • R= 0.082 \frac{atmL}{molK}
  • T= 30 C= 303 K (being 0 C= 273 K)

Replacing in the ideal gas law:

0.888231 atm× 0.165 L = n× 0.082 \frac{atmL}{molK}× 303 K

Solving:

n= (0.888231 atm× 0.165 L)÷ (0.082 \frac{atmL}{molK}× 303 K)

<u><em>n= 0.0059 moles</em></u>

Finally, 0.0059 moles of oxygen is produced at 30 °C and 90 kPa.

<h3>Mass of KClO₃ required</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 3 moles of O₂ are produced by 244.8 grams of KClO₃, 0.0059 moles of O₂ are produced by how much mass of KClO₃?

mass of KClO_{3}= \frac{0.0059 moles of O_{2}x 244.8 grams of KClO_{3}}{3 moles of O_{2}}

<u><em>mass of KClO₃= 0.48144 grams</em></u>

Finally, 0.48144 grams of KClO₃ was decomposed.

Learn more about

the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

ideal gas law:

<u>brainly.com/question/4147359?referrer=searchResults</u>

4 0
2 years ago
What heat is liberated in the formation of 10.0 grams of sulfur hexafluoride, SF6, from the elements sulfur and fluorine?
tensa zangetsu [6.8K]

Answer:

Q=-76.7kJ

Explanation:

Hello,

In this case, for such formation of sulfur hexafluoride, the standard enthalpy of formation is -1220.47 kJ/mol (data extracted from NIST database). Next, we compute the moles in 10.0 grams of sulfur hexafluoride as shown below:

n_{SF_6}=10.0gSF_6*\frac{1molSF_6}{146.06 gSF_6}=0.0685mol

Next, for the given energy, we compute the total heat that is liberated:

Q=-1220.47\frac{kJ}{mol}*0.0685 mol\\\\Q=-76.7kJ

Finally, we conclude such symbol has sense since negative heat is related with liberated heat.

Best regards.

8 0
3 years ago
The Tisza-Danube cyanide spill is an example of all of the following except a general pollution. b accidental pollution. c point
Serjik [45]

Answer: Its B!

Explanation:

4 0
3 years ago
Analysis of a compound of sulfur, oxygen and fluorine showed that it is 31.42% S and 31.35% O, with F accounting for the remaind
Leya [2.2K]

Answer:

The molecular formula is  SO2F2

Explanation:

Step 1: Data given

Suppose the mass of compound = 100 grams

The compound contains:

31.42 % S = 31.42 grams S

31.35 % O = 31.35 grams O

100 - 31.42 - 31.35 = 37.23 F

Molar mass of S = 32.065 g/mol

Molar mass F = 19.00 g/mol

Molar mass O = 16.00 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles S = 31.42 grams / 32.065 g/mol

Moles S = 0.9799 moles

Moles 0 = 31.35 grams / 16.00 g/mol

Moles 0 = 1.959 moles

Moles F = 37.23 grams / 19.00 g/mol

Moles F = 1.959 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.9799 / 0.9799 = 1

F: 1.959/ 0.9799 = 2

O : 1.959 / 0.9799 = 2

The empirical formula is SO2F2

This formula has a molecular mass of 102.06 g/mol

This means the empirical formula is also the molecular formula : SO2F2

3 0
3 years ago
9. During an experiment the students prepared three mixtures A)Starch in water B) Sodium chloride solution C) Tincture of Iodine
il63 [147K]

Answer:

See explanation

Explanation:

Tyndall effect refers to the scattering of light in a solution. Tyndall effect occurs when the size of particles in the solution exceeds 1 nm in diameter. Such solutions are actually called false solutions.

In tincture of iodine, the size of particles in solution is less than 1 nm in diameter hence the solution does not exhibit Tyndall effect. Hence, tincture of iodine is a true solution.

Therefore, if the size of particles in solution exceeded 1nm in diameter, Tyndall effect is observed.

7 0
3 years ago
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