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3 years ago

QUestion 6 and 7

2 answers:

8 0

#6: The living organisms in an ecosystem can be divided into three categories: producers, consumers and decomposers. They are all important parts of an ecosystem.

#7: Nature has its own recycling system: a group of organisms called decomposers. Decomposers feed on dead things: dead plant materials such as leaf litter and wood, animal carcasses, and feces. They perform a valuable service as Earth's cleanup crew

HOPE THIS HELPS

Allisa [31]3 years ago

3 0

#7 decompsers eat on the dead. without decomposers dead leaves, dead animals and dead insects would pile up everywhere

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2 years ago

How many formula units make up 19.0 g of magnesium chloride (MgCl2)?

professor190 [17]

To convert the mass of a compound to formula units, the conversion factor is Avogadro's number, 6.022 x10^23 formula units/ mol and its molar mass. In this case, we are given with 19.0 grams of magnesium chloride which has a mass of 95.21 g/mol. Hence the answer is 1.20 x 10^23 formula units.

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3 years ago

Which compound contains both ionic and covalent

musickatia [10]

Answer:

(3) NaNO₃

Step-by-step explanation:

Sodium nitrate has ionic bonds, because it consists of Na⁺ and NO₃⁻ ions.

However, the nitrate ions have covalent bonds between the O atoms and the central N atoms.

(1) and (2) are wrong. Both N₂O₅ and HCl consist of nonmetals, so they are covalent compounds.

(4) is wrong. NaCl has only ionic bonds between the Na⁺ and Cl⁻ ions

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3 years ago

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BabaBlast [244]

Answers are at the pdf.

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3 years ago

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A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

Where;

$\frac{f_2}{f_1}$ is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356$

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

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3 years ago