All categories

4 years ago

What decomposers are in a estuary ?

Chemistry

1 answer:

belka [17]4 years ago

7 0

They are Common Soil Bacteria Nematodes.

Hope this helps:)

You might be interested in

Nuclear reactions involves

Finger [1]

Nuclear reaction involves two reacting particles a heavy target nucleus and a light bombarding particle and produces two new particles a heavier product nucleus and a lighter ejected particle.

6 0

3 years ago

Which part of the mantle is still a solid but flows like a thick, heavy liquid?

marusya05 [52]

Hello!

Your answer is A, asthenosphere

The asthenosphere is a part of the mantle. It helps move the plates in the Earth.

It is below the lithosphere, between 80 and 200 km below the surface.

Therfore, the asthenosphere is the part of the mantle that is still a solid but flows like a thick, heavy liquid.

Hope this helps!

Have a great day!

3 0

4 years ago

Read 2 more answers

What is △n for the following equation in relating Kc to Kp?

Nimfa-mama [501]

Answer:

-1

Explanation:

The relation between Kp and Kc is given below:

$K_p= K_c\times (RT)^{\Delta n}$

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant , 0.082057 L atm.mol⁻¹K⁻¹

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

$2Na_{(s)}+2H_2O_{(l)}\rightleftharpoons 2NaOH_{(aq)}+2H_2_{(g)}$

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants) = (2+1)-(2+2) = -1

7 0

4 years ago

You are given a sulfuric acid solution of unknown concentration. You dispense 10.00 mL of the unknown solution into an Erlenmeye

dmitriy555 [2]

Answer:

"0.053457 M" of sulfuric acid.

Explanation:

The given values are:

$V$ = 10 mL solution

$V_{added}$ = 12.20 mL

$V_{total}$ = 22.20 mL

then,

M 0.103 M of NaOH,

$V_{rinsed}$ = experiment will not be affected

$V_{total \ base}$ = 10.38 mL

Now,

⇒ mol of NAOH = MV

= $0.103\times 10.38$

= $1.06914 \ m$

Whether Sulfuric acid, then

⇒ $H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O$

⇒ $mol \ of \ acid =\frac{1}{2}\times \ mol \ of \ base$

⇒ $1.06914 \ m \ mol \ of \ base = \frac{1}{2}\times 1.06914 = 0.53457 \ m \ mol \ of \ acid$

Before any dilution:

$V_{sample} = 10 \ mL$

⇒ $M \ acid = \frac{m \ mol}{V}$

$=\frac{ 0.53457 }{10}$

$=0.053457 \ M$ (Sulfuric acid)

6 0

3 years ago

Scientists observe patterns in how molecules interact with each other. At the same temperature, for example, liquids have more a

Alla [95]

Answer:

Hello

I am Parth

Can we became friends

and ur intro plz

7 0

3 years ago