Please help! This is 8th grade science btw

Answer:

group 2

Explanation:

Answer:

this may help

Explanation:

in primary succession, newly exposed or newly formed rock is colonized by living things for the first time. In secondary succession, an area that was previously occupied by living things is disturbed, then re-colonized following the disturbance.

Animals can be present during the start of secondary succession, but they are never present during the start of primary succession is a difference between animal populations in primary succession and secondary succession. Therefore, option A is correct.

When a new piece of land is produced or is first exposed, primary succession occurs. This may occur, for ex., when lava solidifies and forms new rocks or when a glacier recedes and reveals bare rocks. In the initial succession.

Primary succession takes place in a habitat that has never supported life. In a region that was formerly populated but had a disturbance, such a wildfire, secondary succession happens. In this case, dirt would still exist.

In primary succession, living organisms first occupy newly exposed or newly created rock. In secondary succession, a previously inhabited region is disrupted, then recolonized as a result of the disturbance.

Thus, option A is correct.

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Answer:

2-0-1, 1-1-1, 0-1-2, but they must all equal 3 electrons in total

Explanation:

Answer:

there is nothing there?

Explanation:

Answer:

A (Hypothesis) is scientific knowledge published through direct observation and remains constant scientific knowledge can change when scientists (generate hypothesis)

The statements which describes elements undergoing transmutation are the number of nucleons is the same before and after and the parent element is different from the daughter element.

It is defined as a substance which cannot be broken down further into any other substance. Each element is made up of its own type of atom. Due to this reason all elements are different from one another.

Elements can be classified as metals and non-metals. Metals are shiny and conduct electricity and are all solids at room temperature except mercury. Non-metals do not conduct electricity and are mostly gases at room temperature except carbon and sulfur.

The number of protons in the nucleus is the defining property of an element and is related to the atomic number.All atoms with same atomic number are atoms of same element.

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The number of moles of Hydrogen​ : 1.993

Given

1.20 x 10²⁴ hydrogen atoms

Required

The number of moles

Solution

The mole is the number of particles(molecules, atoms, ions) contained in a substance  

1 mol = 6.02.10²³ particles

Can be formulated  :

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

moles of Hydrogen :

[tex]\tt n=\dfrac{N}{No}\\\\n=\dfrac{1.2\times 10^{24}}{6.02\times 10^{23}}\\\\n=1.993~moles[/tex]

Answer:

[Xe] 4f145d106s1

Explanation:

Answer:

the last one

Explanation:

Answer:

D

Explanation:

Answer:

Carbon, in all its states of matter.

Explanation:

Answer:

Closed

Explanation:

Answer:

The Himalayas

Explanation:

Answer:

Base

Explanation:

Answer:

Q 1 oxygen 2 6

Q2 D

Q3 B

4 Mg

5 B

Explanation:

Answer:

A 4 half-life

B. 0.05 mg

Explanation:

A. Determination of the number of half-lives after 2.8×10⁹ years.

From the question given above,

7×10⁸ years = 1 half life

Therefore

2.8×10⁹ years = 2.8×10⁹ years × 1 half life / 7×10⁸ years

2.8×10⁹ years = 4 half life

Thus, the sample went through 4 half-lives at the end of 2.8×10⁹ years.

B. Determination of the amount of the sample remaining after 2.8×10⁹ years.

Original amount (N₀) = 0.74 mg

half life (t½) = 7×10⁸ years

Time (t) = 2.8×10⁹ years

Amount remaining (N) =?

Next, we shall determine the rate of disintegration. This can be obtained as follow:

half life (t½) = 7×10⁸ years

Decay constant (K) =?

K = 0.693 / t½

K = 0.693 / 7×10⁸

K = 9.9×10¯¹⁰ /year

Finally, we shall determine the amount remaining as follow:

Original amount (N₀) = 0.74 mg

Time (t) = 2.8×10⁹ years

Decay constant (K) = 9.9×10¯¹⁰ /year

Amount remaining (N) =?

Log (N₀/N) = kt / 2.303

Log (0.74/N) = 9.9×10¯¹⁰×2.8×10⁹ /2.303

Log (0.74/N) = 2.772 / 2.303

Log (0.74/N) = 1.2036

Take the antilog of 1.2036

0.74/N = antilog (1.2036)

0.74 / N = 15.98

Cross multiply

0.74 = N × 15.98

Divide both side by 15.98

N = 0.74 / 15.98

N = 0.05 mg

Thus, 0.05 mg of the sample will remain after 2.8×10⁹ years

The amount of uranium sample remained after 4 cycles in [tex]\rm 2.8\;\times\;10^9[/tex] years has been 0.04625 mg.

The half-life can be described as the time required by the element to reduce to its half concentration from the initial concentration.

A. The number of half-life cycles can be calculated as:

[tex]\rm 7.0\;\times\;10^8[/tex] = 1 cycle

[tex]\rm 2.8\;\times\;10^9[/tex] = [tex]\rm \dfrac{1}{\rm 7.0\;\times\;10^8}\;\times\;2.8\;\times\;10^9[/tex]

= 4 cycles.

The number of half-life cycles after [tex]\rm 2.8\;\times\;10^9[/tex] years are 4 cycles.

B. The amount of sample remained can be calculated as:

Sample remained = Initial sample [tex]\rm \times\;\dfrac{1}{2}^\frac{time}{Half-life}[/tex]

Sample remained = 0.74 mg [tex]\rm \times\;\dfrac{1}{2}^\frac{2.8\;\times\;10^9}{7.0\;\times\;10^8}[/tex]

Smaple remianed = 0.74 [tex]\rm \times\;\dfrac{1}{2}^4[/tex]

Sample remained = 0.74 [tex]\times[/tex] 0.0625 mg

Sample remained = 0.04625 mg.

The amount of uranium sample remained after 4 cycles in [tex]\rm 2.8\;\times\;10^9[/tex] years has been 0.04625 mg.

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Answer:

it's called an organ system

Answer:

organ system

Explanation:

I Honestly dont know if its correct but worth a shot

Answer:

Isotopes are atoms of the same element that have different numbers of neutrons but the same number of protons and electrons. The difference in the number of neutrons between the various isotopes of an element means that the various isotopes have different masses.

Hope it helped

Answer:

20 g Ag

General Formulas and Concepts:

Chemistry - Stoichiometry

Chemistry - Atomic Structure

Explanation:

Step 1: Define

[RxN]   Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)

[Given]   10 g Cu

Step 2: Identify Conversions

[RxN]   1 mol Cu = 1 mol Ag

Molar Mass of Cu - 63.55 g/mol

Molar Mass of Ag - 197.87 g/mol

Step 3: Stoichiometry

[tex]10 \ g \ Cu(\frac{1 \ mol \ Cu}{63.55 \ g \ Cu})(\frac{1 \ mol \ Ag}{1 \ mol \ Cu} )(\frac{197.87 \ g \ Ag}{1 \ mol \ Ag} )[/tex] = 16.974 g Ag

Step 4: Check

We are given 1 sig fig. Follow sig fig rules and round.

16.974 g Ag ≈ 20 g Ag

Answer:

true

Explanation:

Answer:

False..

Explanation: parasites do not provide you with anything good. The parasite obtains nutrition directly from the body of the host

Answer:c or d

Explanation:

Answer:

yall its b

Explanation: cus i took the tested

Answer:

The answer is number 3

Explanation:

hope it helps

Answer: The volume of this sample at STP is 15.4 L

Explanation:

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas =  210 kPa

[tex]P_2[/tex] = final pressure of gas = 101.3 kPa ( at STP)

[tex]V_1[/tex] = initial volume of gas = 9.50 L

[tex]V_2[/tex] = final volume of gas =  ? ( at STP)

[tex]T_1[/tex] = initial temperature of gas = 350 K

[tex]T_2[/tex] = final temperature of gas = 273 K  ( at STP)

Now put all the given values in the above equation, we get:

[tex]\frac{210\times 9.50}{350}=\frac{101.3\times V_2}{273}[/tex]

[tex]V_2=15.4[/tex]

The volume of this sample at STP is 15.4 L

Answer:

biosphere

Explanation:

Living things need water (hydrosphere), chemicals from the atmosphere, and nutrients gained by eating things in the biosphere.

Answer:

Extrusive and intrusive are the two major types of igneous rocks. On the Earth's surface, extrusive rocks are created from lava, which is magma that has erupted from the underground. Intrusive rocks are formed from magma within the planet's crust that cools and solidifies.

Explanation:

Yeet

Answer:

lithium iodide is formed by sharing of electrons where is Potassium iodide is formed by opposite charges and electrostatic force between them.

Explanation: