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3 years ago

Which 4 elements make up the bulk of living matter?

1 answer:

4 0

"<span>Let's start with the basics: you have probably heard that your body is 90% (or 85% or 95%) water. In actual fact, water is (probably--unless you are obese) the majority of your body weight--but nowhere near 90%. The best estimate I have seen is 57% (Body water). So that puts hydrogen and oxygen in the top four.

As I mentioned, extremely obese people can be less than 50% water. Fat (triglycerides) is composed of carbon, hydrogen, and oxygen. The majority of the mass of a triglyceride is carbon--chemically it is closely related to biodiesel. So that puts carbon in the top four.

Finally, amino acids which make up proteins (which make up muscle) require nitrogen. That makes the top four.: Oxygen, Carbon, Hydrogen, and Nitrogen.

But wait! What about bones? Aren't they calcium and phosphorous? Bones are really heavy.

It turns out that calcium and phosphorous are indeed the next two most abundant (by mass) elements in the body, but where nitrogen is about 3% of your body mass, calcium and phosphorous are 1.5% and 1.2% respectively.

You can find the full breakdown of the elemental composition of your body here: <span>Abundance of the chemical elements"

</span></span>

You might be interested in

Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.45 M of reagent

sladkih [1.3K]

Answer:

$\large\boxed{\large\boxed{0.0014M/s}}$

Explanation:

From the table, first find the order of reaction, then find the rate constant, write the rate equation, and, finally, subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

<u>1. Table</u>

Trial [A] (M) [B] (M) [C] (M) Initial rate (M/s)

1 0.20 0.20 0.20 6.0×10⁻⁵

2 0.20 0.20 0.60 1.8×10⁻⁴

3 0.40 0.20 0.20 2.4×10⁻⁴

4 0.40 0.40 0.20 2.4×10⁻⁴

<u>2. Orders</u>

a) From trials 3 and 4 you learn that the initial concentration of B does not change the change teh rate of the reaction. Hence the order with respect to [B] is 0.

b) From trials 1 and 2 you learn that when the concentration of C is tripled the rate of reaction is also tripled:

0.60 / 0.2 = 3, and

1.8×10⁻⁴ / 6.0×10⁻⁵ = 3

Hence, the order with respect to C is 1.

c) From trials 1 and 3 you get:

0.40/0.2 = 2

2.4×10⁻⁴ / 6.0×10⁻⁵ = 4

Which means that when the concentration of A is doubled, the rate of the reaction is quadruplicated. Hence, the order of the reaction with respect to A is 2.

<u>3. Rate equation</u>

Ther orders are:

$a=2\\\\b=0\\\\c=1$

Hence the rate is:

$rate=k[A]^a{B}^b[C]^c\\ \\ rate=k[A]^2[B]^0[C]^1=k[A]^2C$

<u>4. Rate constant, k</u>

<u />

You can use any trial to find the value of the constant, k

Using trial 1:

$6.0\times 10^{-5}M/s=k(0.20M)^2(0.20M)\\ \\ k=\frac{6.0\times 10^{-5}M/s}{(0.20M)^2(0.20M)}=0.0075M^{-2}s^{-1}$

$rate=k[A]^2C=0.0075[A]^2[C]$

<u>6. Substitute</u>

Subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

$rate=0.0075M^{-2}s^{-1}[A]^2[C]=0.0075M^{-2}s^{-1}[0.45M]^2[0.9M]$

$r=0.00136688M/s\approx 0.0014M/s$

7 0

3 years ago

What stress would shift the equilibrium position of the following system to the right?

yanalaym [24]

Answer:

Decreasing the concentration of N2O3

Explanation:

This is because the products on the right of the reaction occupy more space. One (1) mole of NO and another mole of NO2 will occupy more space than the one (1) mole of N2O3. Therefore decreasing the concentration of N2O3 will shift the reaction to the right because the products will have more space to occupy – hence favoring equilibrium.

4 0

3 years ago

Read 2 more answers

Which group will have an electron configuration that ends in s2?

Step2247 [10]

<h2>Answer: c . Alkaline earth metals (Group 2)</h2>

<h3>Explanation:</h3>

Group 2 metals have 2 electrons in their outer shell. These two electrons are usually found in the s orbital, hence the s².