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Fed [463]
3 years ago
5

The protons of an atom are

Chemistry
2 answers:
Julli [10]3 years ago
8 0
The protons of an atom are P<span>ositively charged and located inside the nucleus.
</span>
Triss [41]3 years ago
4 0
B protons are positive atoms and are inside the nucleus
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In a chemical interaction 10 g of hydrogen chloride is added to 20 g of zinc to form 24 g of zinc chloride and a certain amount
dybincka [34]
6g of hydrogen gas is my answer. I'm sorry if I'm wrong.
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why does a properly adjusted head restraint help prevent head and neck injuries to occupants in rear end collision?
Kazeer [188]
If it's properly adjusted ur head and nexk will stay in place for no further injuries.It helps so it don't move
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22.4 L is the volume of any gas regardless of atmospheric conditions.<br><br> O True<br><br> O False
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8 0
3 years ago
What is the correct 1 letter code for the amino acid sequence Glutamic Acid-Histadine-Tyrosine Select one: a. E-H-Y b. G-H-T c.
EastWind [94]

Answer:

a. E-H-Y

Explanation:

A group of three nucleotides is called a codon that codes for a specific amino acid in the protein. There are 20 essential amino acids present in human body and are required in the diet.

Each amino acid is given a one-letter code that makes the study of amino acid sequences easy. One letter code for the given amino acid sequence Glutamic Acid-Histidine-Tyrosine is E-H-Y in which E is code for Glutamic Acid, H is a code for Histidine, and Y is a code for Tyrosine.

Hence, the correct answer is "a. E-H-Y".

6 0
3 years ago
What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha
sladkih [1.3K]

Answer:

The mean free path = 2.16*10^-6 m

Explanation:

<u>Given:</u>

Pressure of gas P = 100 kPa

Temperature T = 300 K

collision cross section, σ = 2.0*10^-20 m2

Boltzmann constant, k = 1.38*10^-23 J/K

<u>To determine:</u>

The mean free path, λ

<u>Calculation:</u>

The mean free path is related to the collision cross section by the following equation:

\lambda =\frac{1}{n\sigma }------(1)

where n = number density

n = \frac{P}{kT}-----(2)

Substituting for P, k and T in equation (2) gives:

n = \frac{100,000 Pa}{1.38*10^{-23} J/K*300K} =2.42*10^{25}\  m^{3}

Next, substituting for n and σ in equation (1) gives:

\lambda =\frac{1}{2.42*10^{25}m^{-3}* .0*10^{-20}m^{2}}=2.1*10^{-6}m

6 0
3 years ago
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