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marysya [2.9K]
3 years ago
14

What is the meaning of the word organism

Chemistry
2 answers:
Vlada [557]3 years ago
8 0

Answer:

Organism is a creature such as plant ,animals and living things or a single-celled life form

Explanation:

The living creature is compared with the interdependent part of the organism. Some of the Example for the organism is dog, bacteria or unicellular animal-like amoeba. This organism can be classified by naming and grouping. The biology name of an organism is called Taxonomy. Classification of an organism help in exploring the unknown potential of many helpful organisms.Due to their classification we can restore the ecosystem, which is imbalanced today.

Lady_Fox [76]3 years ago
5 0

Answer:

B) any complex thing with properties normally associated with living things

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Answer: Quantitative data

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How the ph values can be used to compare the strength of hydrochloric acid and ethanoic acid?
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3 years ago
The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

8 0
3 years ago
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Answer:

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8 0
3 years ago
Read 2 more answers
What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)
rewona [7]

Answer:

Ka = 4.04 \times 10^{-11}

Explanation:

Initial concentration of weak acid = 4.5 \times 10^{-4}\ M

pH = 6.87

pH = -log[H^+]

[H^+]=10^{-pH}

[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M

HA dissociated as:

HA \leftrightharpoons H^+ + A^{-}

(0.00045 - x)    x     x

[HA] at equilibrium = (0.00045 - x) M

x = 1.35 \times 10^{-7}\ M

Ka = \frac{[H^+][A^{-}]}{[HA]}

Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}

0.000000135 <<< 0.00045

Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}

5 0
3 years ago
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