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3 years ago

8

Which of the following is not an element that makes up all living organisms?

1 answer:

Bess [88]3 years ago

8 0

Answer: Rubidium is correct

Explanation: The six most common elements in living things are carbon, hydrogen, oxygen, nitrogen, phosphorus, and sulfur. Atoms of these elements combine and form thousands of large molecules. These large molecules make up the structures of cells and carry out many processes essential to life.

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True or False <br> the way energy flows through an ecosystem can be shown through a food web

spayn [35]

True energy passes from animal to animal or plant to plant

3 0

3 years ago

Read 2 more answers

The activation energy of a reaction is 56.9 kj/mol and the frequency factor is 1.5×1011/s. Part a calculate the rate constant of

Vanyuwa [196]

We will use Arrehenius equation

lnK = lnA -( Ea / RT)

R = gas constant = 8.314 J / mol K

T = temperature = 25 C = 298 K

A = frequency factor

ln A = ln (1.5×10 ^11) = 25.73

Ea = activation energy = 56.9 kj/mol = 56900 J / mol

lnK = 25.73 - (56900 / 8.314 X 298) = 2.76

Taking antilog

K = 15.8

5 0

4 years ago

Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y

olya-2409 [2.1K]

Answer:

0.9715 Fraction of Pu-239 will be remain after 1000 years.

Explanation:

$\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

$A=A_o\times e^{-\lambda t}$

Where:

$\lambda$ = decay constant

$A_o$ =concentration left after time t

$t_{\frac{1}{2}}$ = Half life of the sample

Half life of Pu-239 = $t_{\frac{1}{2}}=24,000 y$ [

$\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]$

Let us say amount present of Pu-239 today = $A_o=x$

A = ?

$A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}$

$A=0.9715\times x$

$\frac{A}{A_0}=\frac{A}{x}=0.9715$

0.9715 Fraction of Pu-239 will be remain after 1000 years.

7 0

3 years ago

Boron sulfide, B2S3(s), reacts violently with water to form dissolved boric acid (H3BO3) and hydrogen sulfide gas.Express your a

ioda

Answer:

Explanation:

From the statement of the problem,

B₂S₃ $_{s}$ + H₂O $_{l}$ → H₃BO₃ $_{aq}$ + H₂S $_{g}$

B₂S₃ + H₂O → H₃BO₃ + H₂S

We that the above expression does not conform with the law of conservation of mass:

To obey the law, we need to derive a balanced reaction equation:

Let us use the mathematical method to obtain a balanced equation.

let the balanced equation be:

aB₂S₃ + bH₂O → cH₃BO₃ + dH₂S

where a, b, c and d will make the equation balanced.

Conservating B: 2a = c

S: 3a = d

H: 2b = 3c + 2d

O: b = 3c

if a = 1,

c = 2,

b = 6,

2d = 2(6) - 3(2) = 6, d = 3

Now we can input this into our equation:

B₂S₃ + 6H₂O → 2H₃BO₃ + 3H₂S

B₂S₃ $_{s}$ + 6H₂O $_{l}$ → 2H₃BO₃ $_{aq}$ + 3H₂S $_{g}$

4 0

3 years ago

describe the placement of the crucible lid on the crucible when heating the magnesium. Why is it important that this be done cor

Andreas93 [3]

What
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4 years ago