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Dafna11 [192]
3 years ago
13

what pressure must be applied to compress a gas from 85 L at 3.5 atm to 15 L? a-17.3atm b-24.7atm c-19.8atm d-21.4atm

Chemistry
1 answer:
pychu [463]3 years ago
5 0
A... I think ., not sure
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Please help! 20 Points!
Pani-rosa [81]

Answer: YOU MAY FIND THE ANSWER IF YOU SEARCH THE WEB.

Explanation:

7 0
3 years ago
What is the concentration of H3O+ ions in saliva if [OH-] = 4.22 x 10-10 M? Provide the pH and the classification of this sample
OLEGan [10]

pH=4.625

The classification of this sample of saliva : acid

<h3>Further explanation</h3>

The water equilibrium constant (Kw) is the product of concentration

the ions:

Kw = [H₃O⁺] [OH⁻]

Kw value at 25° C = 10⁻¹⁴

It is known [OH-] =  4.22 x 10⁻¹⁰ M

then the concentration of H₃O⁺:

\tt 10^{-14}=4.22\times 10^{-10}\times [H_3O^+]\\\\(H_3O^+]=\dfrac{10^{-14}}{4.22\times 10^{-10}}=2.37\times 10^{-5}

pH=-log[H₃O⁺]\tt pH=5-log~2.37=4.625

Saliva⇒acid(pH<7)

6 0
3 years ago
How many grams of K2Cr207 are in 250. mL of 0.500M of K2Cr207:
goldenfox [79]

<u>Analysing the Question:</u>

We are given a 250 mL solution of 0.5M K₂Cr₂O₇

Which means that we have:

0.5 Mole in 1L of the solution

0.125 moles in 250 mL of the solution      <em>[dividing both the numbers by 4]</em>

<em />

<u>Mass of K₂Cr₂O₇ in the given solution:</u>

Molar mass of K₂Cr₂O₇(Potassium Dichromate) = 194 g/mol

<em>we know that we have 0.125 moles in the 250 mL solution provided</em>

Mass = Number of moles * Molar mass

Mass = 0.125 * 194

Mass = 36.75 grams

7 0
3 years ago
Notice
mixer [17]

Answer:

gahwidsuacsgsuacayau1joagavahiq8wtw8quavakiafabajozyavqhaigavayquata

Explanation:

vahaiqgahiavavqugafayqigqvsbjsiagwyeiwvvs

3 0
3 years ago
How much heat is required to increase the temperature of 100.0 grams of iron from 15.0oC to 40.2oC? (The specific heat of iron i
Ksenya-84 [330]
Answer is: 1160 J of heat Is required to increase the temperature.
m(Fe) = 100 g.
∆T = 40,2 - 15 = 25,2°C.
C(Fe) = 0,46 J/g•°C.
Q = m(Fe) • C • ∆T.
Q = 100 g • 0,46 J/g•°C • 25,2°C
Q = 1160 J.
C - specific heat.

3 0
3 years ago
Read 2 more answers
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