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3 years ago

Which is a nonliving factor that defines a spruce-fir forest ecosystem?

Chemistry

2 answers:

Zanzabum3 years ago

5 0

Soil because trees, bacteria, and mammals are living organisms/things and soil is not.

VashaNatasha [74]3 years ago

4 0

A is the right answer

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Cans someone help me lol ;) WILL MARK BRAINLIEST!!

lidiya [134]

Answer:

The answer to your question is:

Explanation:

1.- Noble gases are special because they do not react, they are stable because they have 8 electrons in their outermost shell.

2.- Hellium

Neon

Argon

Krypton

Xenon

Radon

3.- Magnesium

Calcium

4.- Magnesium

5.- VII A Halogen

6.- III A, IVA, VA, VIA

7.- Nitrogen

8.- Categories of the periodic table

Metals Metalloids Non-metals

Group IA and IIA Groups IIIA to VIA Groups IVA to VIIA

Groups B's

9.- Left side

10.- right side

11.- non metal / left

3 0

3 years ago

A rigid, closed vessel of volume V =648 liters maintained at constant temperature T=365 K is loaded with 112 mol of n-hexane, 15

Marrrta [24]

Answer:

P = 2.92 atm

Explanation:

With the three assumptions in mind, the system consists of:

A liquid phase containing n-hexane and n-heptane, and

A gaseous phase containing n-hexane vapor, n-heptane vapor, and nitrogen gas.

First we use PV=nRT to calculate the moles of n-hexane and n-heptane in the gaseous phase:

n-hexane:

P = 0.199 MPa ⇒ 0.199 * 1.869 = 1.964 atm

1.964 atm * 648 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 365 K

n = 42.52 moles

n-heptane:

P = 0.083 MPa ⇒ 0.083 * 1.869 = 0.155 atm

0.155 atm * 648 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 365 K

n = 3.358 moles

So the gaseous phase consists of 42.52 moles of n-hexane, 3.358 moles of n-heptane, and 14 mol of nitrogen.

For the liquid phase, we calculate the remaining moles of n-hexane and n-heptane. Then we convert to liters, using their molar volumes:

n-hexane:

n = 112 mol - 42.52 mol = 69.48 mol

69.48 mol * 0.146 L/mol = 10.14 L

n-heptane:

n = 155 mol - 3.358 mol = 151.642 mol

151.642 mol * 0.162 L/mol = 24.57 L

So the liquid phase occupies (10.14+24.57) = 34.71 L, and contains 69.48 mol of n-hexane and 151.64 mol of n-heptane.

Finally, to calculate the pressure in the vessel, we use PV=nRT:

P = ?

V = 648 - 34.71 = 613.29 L

n = 42.52 mol hexane + 3.35 mol heptane + 14 mol nitrogen = 59.87 mol

T = 365 K

P * 613.29 L = 59.87 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 365 K

P = 2.92 atm

6 0

3 years ago

How can alcohol content be increased

konstantin123 [22]

Answer:

Alcohol is a byproduct of the fermentation process, which takes place when the yeast converts the sugars derived from the grain. Knowing that, you can increase the alcohol by volume (ABV) by increasing the size of the grain bill or increasing the amount of malt extract used.

4 0

2 years ago

This is due today so please help me

Nonamiya [84]

B north to south. Hope this helps

7 0

3 years ago

Read 2 more answers

Limiting Reactant

liraira [26]

Answer:

(1) Cl₂ is the limiting reactant.

(2) 8.18 g

Explanation:

2Na(s) + Cl₂(g) → 2NaCl(s)

First we convert the given masses of reactants into moles, using their respective molar masses:

Na ⇒ 12.0 g ÷ 23 g/mol = 0.522 mol Na

Cl₂ ⇒ 5.00 g ÷ 70.9 g/mol = 0.070 mol Cl₂

0.070 moles of Cl₂ would react completely with (2 * 0.070) 0.14 moles of Na. There are more Na moles than that, so Na is the reactant in excess while Cl₂ is the limiting reactant.

Then we calculate how many moles of NaCl are formed, using the limiting reactant:

0.070 mol Cl₂ * $\frac{2molNaCl}{1molCl_2}$ = 0.14 mol NaCl

Finally we convert NaCl moles into grams:

0.14 mol NaCl * 58.44 g/mol = 8.18 g

3 0

3 years ago