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aleksandr82 [10.1K]
3 years ago
9

How do scientsts determine is a newly found thing is dead or alive.

Chemistry
1 answer:
butalik [34]3 years ago
4 0

The substance is dead or alive is determined by response or consciousness of the substance to external stimulus.

<h3><u>Explanation</u>:</h3>

The living organisms do have several characteristics. They are growth, reproduction, consciousness etc. But among them, consciousness is the only feature that's confined within only the living world. No non living things have this feature.

So if the scientist can find that the substance found have some response to external stimulus, then it does have life. And if it doesn't response to any sort of external stimulus, then it doesn't have life.

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What was the hiker's average velocity during part C of the hike?
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A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles o
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<u>Answer:</u> The value of K_{eq} is 0.044

<u>Explanation:</u>

We are given:

Initial moles of methane = 4.10\times 10^{-2}mol=0.0410moles

Initial moles of carbon tetrachloride = 6.51\times 10^{-2}mol=0.0651moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of methane = \frac{0.0410}{1.00}=0.0410M

Concentration of carbon tetrachloride = \frac{0.0651}{1.00}=0.0651M

The given chemical equation follows:

                    CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)

<u>Initial:</u>          0.0410    0.0651

<u>At eqllm:</u>     0.0410-x   0.0651-x       2x

We are given:

Equilibrium concentration of carbon tetrachloride = 6.02\times 10^{-2}M=0.0602M

Evaluating the value of 'x', we get:

\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M

Now, equilibrium concentration of methane = 0.0410-x=[0.0410-0.0049]=0.0361M

Equilibrium concentration of CH_2Cl_2=2x=[2\times 0.0049]=0.0098M

The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}

Putting values in above expression, we get:

K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044

Hence, the value of K_{eq} is 0.044

8 0
3 years ago
Iron fluoride (FeF2) dissociates according to the following equation:
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Answer:

S = 0.788 g/L

Explanation:

The solubility product (Kps) is an equilibrium solubization constant, which can be calculated by the equation:

Kps = \frac{[product]^x}{[reagent]^y}

Where x and y are the stoichiometric coefficients of the product and the reagent, respectively. Because of the aggregation form, the concentration of solids is always equal to 1 for use in this equation.

Analyzing the equation, we see that for 1 mol of Fe^{+2} is necessary 2 mols of F^-, so if we call "x" the molar concentration of Fe^2, for F^- we will have 2x, so:

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Fe = 55.8 g/mol

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FeF2 = Fe + 2xF = 55.8 + 2x19 = 93.8 g/mol

So,

[tex]S = 8.4x10^{-3}x93.8

S = 0.788 g/L

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