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2 years ago

11

identify the part of the slow carbon cycle in whitch the total amount of carbon is most likely decreasing and explain why this d

ecrease occurs

1 answer:

Westkost [7]2 years ago

3 0

Answer:

i just had gthe sme question looked it up and got it WRONG

Explanation:

sorry i ont have a clue

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Iodine-131 has a half-life of 8 days. If the amount of iodine-131 in the orginial sample is 8 g, how much iodine-131 will remain

hram777 [196]

1/32

Explanation it stated with half on 8 days then that means you divide 24 and 8 so its 3 and you have to multiply ✖ ½

7 0

2 years ago

Which example is a trace fossil? {A} dinosaur egg {B} dinosaur footprint {C} dinosaur bone {D} shark tooth

viktelen [127]

dinosaur footprint

Explanation:

A dinosaur footprint is an example of a trace fossil. A trace fossil is a type of fossil that shows the activities of organisms that lived in the past.

Fossils are the preserved remains of organisms that lived several years ago.
Fossils are usually found in sedimentary rocks and thick layers of ice in temperate and polar regions.
Body fossils are the remains of the body parts of an organism that has been preserved. They can be skeletal parts, teeth, eggs e.t.c
A trace fossil shows the preserved remains of the activities of an organism.
They can be fingerprints, burrows and borings, feccal pellets e.t.c

Learn more":

fossils and evolution brainly.com/question/12790206

#learnwithBrainly

5 0

2 years ago

Read 2 more answers

What’s the distance from the nucleus of an atom to its theoretical outer edge of electron orbits

ollegr [7]

Answer:

The range of atoms = (30-300 pm) depending upon the element

Explanation:

The Atomic radii of the atom is the distance from the center of the circle to the outermost orbital.

The center of the circle is the nucleus and the radii is the outermost boundary.

The actual size of the atom is decided on the basis of the Zeff . Also known as <em>effective nuclear charge.</em>

<em>Zeff: It is the net positive charge felt by the outermost electron by the nucleus.</em>

<em>The value of Zeff depends upon the shielding constant. More the shielding less will be the Zeff . Hence the size of the atom increases.</em>

Due to shielding the outermost electrons feel less pull of nucleus.

<em>The greater the Zeff , the smaller the radius of the atom.</em>

The formula used to calculate the atomic mass is :

$r_{n}=\frac{52.9n^{2}}{Z}$ pm

Here "pm"= picometers

$1 pm = 10^{-12}m$

<u>The size of the smallest atom H-atom = 120 pm</u>

<u>The range of atoms = (30-300 pm)</u>

4 0

2 years ago

Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241

anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

$XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)$ $\Delta H_1=-241.8kJ$

(2) $X(s)+2Cl_2(g)\rightarrow XCl_4(s)$ $\Delta H_2=+461.9kJ$

(3) $\frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)$ $\Delta H_3=-92.3kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $H_2O(g)\rightarrow H_2O(l)$ $\Delta H_5=-44.0kJ$

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)$ $\Delta H_1=2\times 241.8kJ=483.6kJ$

(2) $XCl_4(s)\rightarrow X(s)+2Cl_2(g)$ $\Delta H_2=-461.9kJ$

(3) $2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)$ $\Delta H_3=4\times -92.3kJ=-369.2kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $2H_2O(l)\rightarrow 2H_2O(g)$ $\Delta H_5=2\times 44.0kJ=88.0kJ$

The expression for enthalpy of main reaction will be:

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5$

$\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)$

$\Delta H=-1048.6kJ$

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

4 0

2 years ago

Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings: 175, 104, 164, 193, 13

Mila [183]

Answer:

Following are the solution to these question:

Explanation:

Calculating the mean:

$\bar{x}=\frac{175+104+164+193+131+189+155+133+151+176}{10}\\\\$

$=\frac{1571}{10}\\\\=157.1$

Calculating the standardn:

$\sigma=\sqrt{\frac{\Sigma(x_i-\bar{x})^2}{n-1}}\\\\$

Please find the correct equation in the attached file.

$=28.195$

For point a:

$=3s+yblank \\\\=3 \times 28.195+50\\\\=84.585+50\\\\=134.585\\$

For point b:

$=3 \ \frac{s}{m}\\\\ = \frac{(3 \times 28.195)}{1.75 \times 10^9 \ M^{-1}}\\\\= 4.833 \times 10^{-8} \ M$

For point c:

$= 10 \frac{s}{m} \\\\= \frac{(10 \times 28.195)}{1.75 x 10^9 \ M^{-1}}\\\\ = 1.611 \times 10^{-7}\ M$

It is calculated by using the slope value that is $1.75 \times 10^9 M^{-1}$ . The slope value $1.75 \times 10^9 M^{-1}$ is ambiguous.

7 0

1 year ago