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2 years ago

Which place is a likely location for primary succession

Chemistry

2 answers:

murzikaleks [220]2 years ago

8 0

Answer:

Places where primary succession occurs

include newly exposed rock areas, sand dunes, and

lava flows.

ZanzabumX [31]2 years ago

3 0

Primary succession occurs in an area that has not been previously occupied by a community. Places where primary succession occurs include newly exposed rock areas, sand dunes, and lava flows. Simple species that can tolerate the often- harsh environment become established first.

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What role does the wire play in voltaic cell?

Triss [41]

<h3><u>Answer;</u></h3>

a) It allows electrons to flow from the anode to the cathode.

<h3><u>Explanation</u>;</h3>

<em><u>Voltaic cell is an electrochemical cell in which a spontaneous chemical reaction produces the flow of electrons</u></em>.
Electrons are produced by the oxidation reaction occurring at the anode. Electrons flow through the conducting wire from the anode to the cathode. At the cathode these electrons are used to reduce copper(II) ions to copper atoms.
<em><u>A conducting wire or a wire play connects the two electrodes allowing electrons to flow from the anode to the cathode</u></em>.

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2 years ago

Dos moléculas de Clorato de Potasio en estado sólido, al aplicarle calor se descompone en dos moléculas de Cloruro de Potasio en

DedPeter [7]

Answer:

2KClO3 (s) -------> 2KCl (s) + 3O2 (g)

Explanation:

Una ecuación de reacción química es una representación simbólica de lo que sucede en un recipiente de reacción.

Al escribir la ecuación de reacción química para una reacción particular, se utilizan los símbolos químicos de todas las especies involucradas.

Por la reacción. en el cual dos moléculas de Clorato de Potasio en estado sólido, al aplicar calor se descompone en dos moléculas de Cloruro de Potasio en estado sólido y tres moléculas diatómicas de Oxígeno en estado gaseoso, la ecuación de reacción se escribe así;

2KClO3 (s) -------> 2KCl (s) + 3O2 (g)

La regla para escribir ecuaciones de reacción química balanceada es que el número de átomos de cada elemento en el lado derecho de la ecuación de reacción debe ser el mismo que el número b de átomos del mismo elemento en el lado izquierdo de la ecuación de reacción.

Si realizamos un conteo de átomos simple en ambos lados de la ecuación de reacción, la regla se cumple

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2 years ago

Chemistryyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

mrs_skeptik [129]

Water is your answer!!!!!!!!!

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2 years ago

What is the mass of 3.77mol of K3N?

AleksAgata [21]

<h3>Answer:</h3>

495 g K₃N

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.77 mol K₃N

<u>Step 2: Identify Conversions</u>

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.77 \ mol \ K_3N(\frac{131.31 \ g \ K_3N}{1 \ mol \ K_3N})$
Multiply/Divide: $\displaystyle 495.039 \ g \ K_3N$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

495.039 g K₃N ≈ 495 g K₃N

5 0

3 years ago

A student is adding DI water to a volumetric flask to make a 50% solution. Unfortunately, he was not paying attention and filled

dalvyx [7]

Answer:

His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.

Explanation:

The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.

Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.

Suppose the mass of the solute is m.

Originally, the density is = $\frac{m}{50}$ $\left(\frac{\text{mass}}{\text{volume}}\right)$