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2 years ago

List the 3 types of organisms and how they each get their energy. * jejdjsjx pls hurryyyl

Chemistry

1 answer:

Helen [10]2 years ago

6 0

Answer:

The three types of organisms in an ecosystem are producers, consumers, and decomposers.

Producers get their energy by using sunlight to make their own food from water and carbon dioxide through photosynthesis.

Consumers get their energy by eating plants and animals for food as they cannot make their own food.

Decomposers get their energy by eating/scavenging and breaking down dead and decaying plant and animal bodies.

Hope that helps.

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Somebody please help me!! What happens when ionic bonds are formed?

xeze [42]

Answer:

The atom that loses the electrons becomes a positively charged ion, while the one that gains them becomes a negatively charged ion

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1 year ago

What are the physical properties of toothpaste

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3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

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2 years ago

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2 years ago

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oee [108]

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2 years ago

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