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Lynna [10]
3 years ago
6

What is the oxidation number of Mn in KMnO₄? a. -7 b. -3 c. 0 d. +3 e. +7

Chemistry
1 answer:
Maurinko [17]3 years ago
4 0
I think the answer is e? That's if o is -2 if not it will be a
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A flashbulb of volume 2.70 mL contains O2(g) at a pressure of 2.30 atm and a temperature of 30.0 °C. How many grams of O2(g) doe
Sholpan [36]
P = 2.30 atm

Volume in liter = 2.70 mL / 1000 => 0.0027 L

Temperature in K = 30.0 + 273 => 303 K

R = 0.082 atm

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number of moles O2 :

P * V = n * R* T

2.30 * 0.0027 = n * 0.082 * 303

0.00621 = n * 24.846

n = 0.00621 / 24.846

n = 0.0002499 moles of O2

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n = m / mm

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m = 0.0002499 * 31.9988

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What titrant will be used to titrate the 0. 02 m hcl phenol red solution?
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The NaOH will be used What titrant  to titrate the 0. 02 m hcl phenol red solution.

Acid-base titrations may be the most typical titrations, although there are numerous more forms as well. Take a look at this illustration where sodium hydroxide is used to titrate a sample of hydrochloric acid (HCl) (NaOH). The titrant (NaOH), which is added gradually throughout the duration of the titration, has been added to the unknown solution.

Titrants are solutions with known concentrations that are added to solutions whose concentrations must be determined. The solution for whom the concentration needs to be determined is known as a titrant as well as analyte.

Therefore, the NaOH will be used as a titrant to titrate the 0. 02 m hcl phenol red solution.

To know more about titrant

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7 0
2 years ago
1.26 * 10^4 + 2.50 * 10^4 in sceintific notation
eduard
Hello! I can help you with this. First, convert them into it’s written out standard form. 10^4 is 10,000. 10,00 * 1.26 is 12,600. 10,000 * 2.5 is 25,000. 12,600 + 25,000 = 37,600 or 3.76 * 10^4 in scientific notation. The answer in scientific notation is 3.76 * 10^4.
8 0
3 years ago
The average distance between nitrogen and oxygen atoms is 115 pm in a compound called nitric oxide. what is this distance in cen
e-lub [12.9K]
<span>pm stands for picometer and picometers are units which can be used to measure really tiny distances. One picometer is equal to 10^{-12} meters. We know that one centimeter is equal to 10^{-2} m so there are 10^2 cm per meter. We can change the distance d = 115 pm to units of centimeters. d = (115 pm) x (10^{-12}m / pm) x (10^2 cm / m) d = 115 x 10^{-10} cm = 1.15 x 10^{-8} cm The distance in centimeters is 1.15 x 10^{-8} cm</span>
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