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3 years ago

What represents a community? pond B a snail snails D snails, guppies and plants

Chemistry

1 answer:

OLga [1]3 years ago

7 0

D snails, guppies and plants. The community should be diverse and have lots of biodiversity.

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In terms of collisions between particles why does the light begin shining brightly and dum overtime

baherus [9]

Answer:

In the case of liquids and solids, the distances between particles are negligible,

Explanation:

3 0

3 years ago

During the process of transpiration, water _______ a plant through the _______.

andrey2020 [161]

Right answer is B . Trust me .

3 0

3 years ago

EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK

mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = $5.0 X 10^{-4} M$

$Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}$

$alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}$

= $2.02 X 10^{-13}$

When,

pH = -log[H+] = 10.15

[H+] = $7.08 X 10^{-11} M$

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = $7.4 X 10^{-7}$ ; Ka6 = $4.3 X 10^-11$

$alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}$

= $5.12 X 10^{-23}$

4 0

3 years ago

Calculate the number of grams of oxygen required to convert 48.0 g of glucose to co2 and h2o.

worty [1.4K]

I believe that the balanced chemical reaction is:

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

So the number of grams of oxygen required is:

mass O2 required = 48 g C6H12O6 * (1 mole C6H12O6 / 180.16 g) * (6 mole O2 / 1 mole C6H12O6) * (32 grams O2 / 1 mole)

mass O2 required = 51.15 grams

4 0

4 years ago

Calculate the ph of a 0.10 m solution of hypochlorous acid, hocl. ka of hocl is 3.5×10−8 at 25 ∘c. express your answer numerical

Viktor [21]

Answer:

4.23.

Explanation:

∵ pH = - log[H⁺].

For weak acids:

∵ [H⁺] = √(ka)(c).

∴ [H⁺] = √(3.5 × 10⁻⁸)(0.10 M) = 5.92 x 10⁻⁵.

∴ pH = - log[H⁺] = - log(5.92 x 10⁻⁵) = 4.2279 ≅ 4.23.

3 0

4 years ago