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3 years ago

6

A student is testing water to know which is best for cleansing purposes with soaps. He would find that the cleansing action of s

oaps is best when he uses water obtained from
A. Rain
B. Tap
C. Hand pump
D. Pond

1 answer:

8_murik_8 [283]3 years ago

6 0

The answer is rain water because it is distilled and distilled water is best for this process

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Ms. Sharkey wanted to see which tennis shoes made her run the fastest. She bought

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Answer:

to compare the data she has got?

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When scientists use one of their five senses to gather information, they are A. making an observation. B. making an inference. C

Goshia [24]

A. making an observation.

Explanation:

When scientists use one of their five sense to gather information, they are simply making an observation.

Most physical properties of matter are studied using sense of taste, of feel, sight, hearing and smell.

Observation is the act of making a measurement or studying.
It helps to investigate particular phenomenon in nature.
Most observations made with our senses are usually qualitative in nature.
They cannot be assigned values like quantitative ones.

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3 years ago

Read 2 more answers

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

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4 years ago

Should the density of a metal depend on its volume?

jekas [21]

Answer:

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Explanation:

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I think it would the hydrosphere

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