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monitta
3 years ago
5

What phases are most common at high temperatures

Chemistry
1 answer:
Nikitich [7]3 years ago
5 0

Answer:

Liquids and gases.

For most substances, when heated to a high enough temperature, the molecules begin moving faster which results in its liquid or gas state.

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The three naturally occurring isotopes of potassium are 39K, 38.963707u; 40K, 39.963999u; and 41K.The percent natural abundances
sweet-ann [11.9K]

Answer:

The isotopic mass of 41K is 40.9574 amu

Explanation:

Step 1: Data given

The isotopes are:  

39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%

40K with an isotopic mass of 39.963999u

41K wit natural abundance of 6.7302 %

Average atomic mass =39.098 amu  

Step 2: Calculate natural abundance of 40 K

100 % - 93.2581 % - 6.7302 %

100 % = 0.0117 %

Step 3: Calculate isotopic mass of 41K

39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302

39.098 = 36.33681 + 0.0046758 +  X * 2.067302

X = 40.9574 amu

The isotopic mass of 41K is 40.9574 amu

8 0
3 years ago
Explain how you can convert from the number of representative particles of a substance to moles of that substance.
Scrat [10]

Answer:

  • the number of representative contain in 1 mol of a substance equals to 6.o2x10^23 particles.
8 0
3 years ago
A concentration cell consisting of two hydrogen electrodes (PH2 = 1 atm), where the cathode is a standard hydrogen electrode and
Lunna [17]

The pH of the unknown solution is 3.07.

<u>Explanation:</u>

<u>1.Find the cell potential as a function of pH</u>

From the Nernst Equation:

Ecell=E∘cell−RT /zF × lnQ

where

R denotes the Universal Gas Constant

T denotes the temperature

z denotes the moles of electrons transferred per mole of hydrogen

F denotes the Faraday constant

Q denotes the reaction quotient

Substitute the values,

E∘cell=0   lnQ=2.303logQ

E0cell=−2.30/RT /zF × log Q

Solving the equation,

<u>2. Find the Q  value</u>

Q=[H+]2prod pH₂, product/ [H+]2reactpH₂, reactant

Q=[H+]^2×1/1×1=[H+]2

Taking the log

logQ= log[H+]^2=2log[H+]=-2pH

From the formula,

Ecell=−2.303RT /zF× logQ

E cell= 2.303 × 8.314 CK mol (inverse)  × 298.15

K × 2pH /2×96 485  C⋅mol

( inverse)

E cell= 0.0592 V × pH

<u>3. Finding the pH value</u>

E cell= 0.0592 V × pH

pH = E cell/ 0.0592 V= 0.182V/ 0.0592V

pH=3.07

The pH of the unknown solution is 3.07.

7 0
3 years ago
What’s the answer ?
Romashka-Z-Leto [24]

Answer:

45.3°C

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 82KPa

Initial temperature (T1) = 26°C

Final pressure (P2) = 87.3KPa.

Final temperature (T2) =.?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

This is illustrated below:

T(K) = T(°C) + 273

Initial temperature (T1) = 26°C

Initial temperature (T1) = 26°C + 273 = 299K.

Step 3:

Determination of the new temperature of the gas. This can be obtained as follow:

P1/T1 = P2/T2

82/299 = 87.3/T2

Cross multiply to express in linear form

82 x T2 = 299 x 87.3

Divide both side by 82

T2 = (299 x 87.3) /82

T2 = 318.3K

Step 4:

Conversion of 318.3K to celsius temperature. This is illustrated below:

T(°C) = T(K) – 273

T(K) = 318.3K

T(°C) = 318.3 – 273

T(°C) = 45.3°C.

Therefore, the new temperature of the gas in th tire is 45.3°C

6 0
3 years ago
How many moles of lithium nitrate will be needed to make 250 grams of lithium sulfate, assuming that you have an excess of lead
frez [133]
The answer would be 313 g
4 0
4 years ago
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