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3 years ago

14

After a day at the beach, bodysurfing and hunting for shark teeth, your parent’s yell that it is time to leave. You grab your sa

ndals and towel and run towards the parking lot. Thirty seconds later you feel like you are walking on hot coals. Explain the sudden change

1 answer:

Lisa [10]3 years ago

5 0

Answer:

There was a change in temperature on the surface you were walking on. It is possible that "you" were walking on cool, wet sand and when your parents said it was time to leave, you ran barefoot towards the parking lot (away from the sand) and got a burning coal sensation on your feet.

Explanation:

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A nuclear fission reaction has mass difference between the products and the reactants of 0.0284 amu. Calculate the

notsponge [240]

Answer:

C) 4.24 x $10^{-12}$

Explanation:

E = Δm $c^{2}$

Δm = 0.0284 x 1.66 x $10^{-27}$ kg = 4.714x $10^{-29}$ kg

putting value in above equation

E = 4.714x $10^{-29}$ kg x (3x $10^{8}$ $)^{2}$ = 4.24 x $10^{-12}$

6 0

2 years ago

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

<u>Answer:</u> The empirical formula for the given compound is $C_3H_6O$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

3 0

3 years ago

ILL MARK BRAINIEST WHOEVER ANSWERS CORRECTLY

natulia [17]

Answer:

I think its A

sorry if im wrong

Explanation:

4 0

2 years ago

What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one a

Leni [432]

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

$\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})$

Where:

Q: is the energy transferred = 5.0 MJ

$k_{B}$ : is the Boltzmann constant = 1.38x10⁻²³ J/K

$T_{i}$ : is the initial temperature = 1000 K

$T_{f}$ : is the final temperature = 500 K

Hence, the entropy change is:

$\Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26}$

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!

7 0

3 years ago

What tool do you use to measure mass?

pshichka [43]

Answer: Balance

Explanation: Mass is the amount of matter contained in a body.

7 0

2 years ago