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3 years ago

Roundworms are what type of biohazards

Chemistry

1 answer:

mr Goodwill [35]3 years ago

7 0

Answer:

Roundworms are parasitic, can cause Toxocariasis

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Which of the following correctly describes a Brønsted-Lowry base?

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If your using Plato the correct answer is (A. It accepts protons)

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3 years ago

an element has a metallic gray appearance. it also has the properties of both a metal and a nonmetal. in which section of the pe

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3 years ago

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4 years ago

Read 2 more answers

g A 0.25 mol sample of HBr is added to a 1.00 L buffer solution consisting of 0.68 M HCN and 0.35 M NaCN. Identify all species t

salantis [7]

Answer:

See explaination for details and diagram

Explanation:

We can say that Chemical reaction is a process in which one or more substances, the reactants, are converted to one or more different substances, the products. Substances are either chemical elements or compounds.

A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products.

See attachment for the buffer of the chemical reaction.

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3 years ago

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is:

$s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}$

$s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}$

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0

3 years ago