Mosquito repellent is sprayed on one arm and the other arm is not sprayed. The number of mosquito bites is counted after 2 hours

What best explains why the box’s final kinetic energy is less than its initial potential energy

ziro4ka [17]

Some potential energy was converted into thermal energy due to friction. The acceleration due to gravity became smaller as the box slid down the ramp.

7 0

3 years ago

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The electrolyte of the following is:

ozzi

An electrolyte is a term used to describe a compound that can dissociate into ions as it is nothing but an ionic compound, a salt made up of a positively charged cation and negatively charged anion.

Here the correct answer is D. Since there are no hydrocarbons or any other organic compound, that do not possess partial let alone full charges, all of them can dissociate in solution to give their ions.

This allows for the solution to be able to conduct electricity.

4 0

3 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago

Why might a scientist repeat an experiment if he/she did not make a mistake in the first one? An experiment should be repeated t

makvit [3.9K]

The answer is C. To ensure the results are accurate.

5 0

3 years ago

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HELP!!!! The freezing of methane is an exothermic change. What best describes the temperature conditions that are likely to make

vazorg [7]

<u>Answer:</u> The correct statement is low temperature only, because entropy decreases during freezing.

<u>Explanation:</u>

The relationship between Gibb's free energy, enthalpy, entropy and temperature is given by the equation:

$\Delta G=\Delta H-T\Delta S$

Where,

$\Delta G$ = change in Gibb's free energy

$\Delta H$ = change in enthalpy

T = temperature

$\Delta S$ = change in entropy

It is given that freezing of methane is taking place, which means that entropy is decreasing and $Delta S$ is becoming negative. It is also given that the reaction is an exothermic reaction, this means that the $\Delta H$ is also negative.

For a reaction to be spontaneous, $\Delta G$ must be negative.

$-ve=-ve-[T(-ve)]\\\\-ve=-ve+T$

From above equations, it is visible that $\Delta G$ will be negative only when the temperature will be low.

Hence, the correct statement is low temperature only, because entropy decreases during freezing.

8 0

3 years ago