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valkas [14]
2 years ago
14

Skeletal muscle is controlled by the organism.

Chemistry
2 answers:
damaskus [11]2 years ago
8 0

Answer:C: Some of the muscle attached to the skeleton is voluntary and may be used for movement.

Explanation:EDGE2021

rosijanka [135]2 years ago
4 0

Some of the muscle attached to the skeleton is voluntary and may be used for movement.

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PLS HELPPP!!!!
prisoha [69]

Answer:

D.

Explanation:

Bohr's model represented electrons in their respective e⁻ shells. Only D fits the description.

5 0
3 years ago
Rank the elements b, be, n, c, and o in order of increasing first ionization energy. (use the appropriate <, =, or > symbo
Dafna11 [192]
Be, B, C, N (going from smallest to biggest)
6 0
3 years ago
Could Rutherford make any conclusions about electrons based on the result of experiment?
MatroZZZ [7]

Answer:

They are so small that they barely make up any of the mass of the atom, and they are so miniscule that during Rutherford 's Gold foil experiment ,they didn't even react with the alpha particles. They circle the nucleus on a ring. shown through Rutherford Atomic Model

hope it helps you

8 0
2 years ago
Assume that you have developed a tracer with first-order kinetics and k = 0.1 s − 1 . For any given mass, how long will it take
anygoal [31]

Answer: It will take 6.93 sec for tracer concentration to drop by 50% and 13.9 sec for tracer concentration to drop by 75%

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  =0.1s^{-1}

t = age of sample  

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  

a) for tracer concentration to drop by 50%

a - x = amount left after decay process   = 50

t=\frac{2.303}{k}\log\frac{100}{50}

t=\frac{0.693}{0.1}

t=6.93sec

It will take 6.93 sec for tracer concentration to drop by 50%

b) for  tracer concentration to drop by 75 %

a - x = amount left after decay process   = 25

t=\frac{2.303}{k}\log\frac{100}{100-75}

t=\frac{2.303}{0.1}\times \log\frac{100}{25}

t=13.9sec

It will take 13.9 sec for tracer concentration to drop by 75%.

5 0
3 years ago
What's the electron configuration of Zr (AN 40)​
nekit [7.7K]
The electron configuration of Zr (AN 40) is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d2 5s2
3 0
2 years ago
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