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enot [183]
3 years ago
6

Garfield (weighing 24 lbs) took a flight from the earth to the moon (where the gravity is 1/6 the gravity here on earth) on the

space shuttle. As usual, he stuffed himself with lasagna during the entire flight and napped when he wasn't eating. Much to his delight when he got to the moon he found he weighed only 6 lbs. He immediately proclaimed a quick weight loss diet.
What is/are the REASONING to support Garfield's claim to a quick weight loss diet?

i need an answer asap please
Chemistry
1 answer:
Nana76 [90]3 years ago
3 0

Answer:

Reason: Gravitational pull of the Moon

Explanation:

Weight of Garfield is 24 lbs on the Earth. He took a flight from the Earth to the Moon where the gravity is 1/6 of the gravity on the Earth.

But when reaches the Moon he found he weighed only 6 lbs. It is due to the gravitational pull of the Moon.

Mass of an object reamins the same at every location while its weight changes due to the change in acceleration due to gravity. It is equal (on Moon) to 1/6 the of acceleration due to gravity on the earth.

You might be interested in
1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium
lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

5 0
3 years ago
Give an example of a non-biological chemical reaction that involves carbon. Name the process that converts a solid compound into
xxMikexx [17]

Answer:

The process that converts a solid mcompound into a gas is sublimation.

E.g- dry ice, solid iodine and ammonium salts

when the above solid are heated, only particles which are found on the surface of the solid gain enough energy and break all forces of attraction and form a gas.

Explanation:

6 0
2 years ago
Who first established the heliocentric view
Alexus [3.1K]
This theory was created by Nicoulas Copernicus
4 0
3 years ago
A balloon is filled with helium at sea level. The volume was measured to be 3.1 liters at a pressure of 0.97 atm. If the balloon
irina [24]
<span>4.9 L would be the answer to this question :)</span>
3 0
3 years ago
Consider the reaction of peroxydisulfate ion (S2O2−8) with iodide ion (I−) in aqueous solution: S2O2−8(aq)+3I−(aq)→2SO2−4(aq)+I−
kolbaska11 [484]

Answer:

r = 3.61x10^{-6} M/s

Explanation:

The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.

r = k.[S2O2^{-8} ]^{x} x [I^{-} ]^{y}

K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :

\frac{r1}{r2} = \frac{0.018^{x} x0.036^{y} }{0.027^xx0.036^y}

\frac{2.6x10^{-6}}{3.9x10^{-6}} = (\frac{0.018}{0.027})^xx(\frac{0.036}{0.036})^y

0.67 = 0.67^x

x = 1

Now, to find the coefficient y let's do the same for the experiments 1 and 3:

\frac{r1}{r3} = \frac{0.018x0.036^y}{0.036x0.054^y}

\frac{2.6x10^{-6}}{7.8x10^{-6}} = (\frac{0.018}{0.036})x(\frac{0.036}{0.054})^y

0.33 = 0.5x 0.67^y

0.67 = 0.67^y

y = 1

Now, we need to calculate the constant k in whatever experiment. Using the first :

2.6x10^{-6} = kx0.018x0.036kx6.48x10^{-4} = 2.6x10^{-6}

k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]

Using the data given,

r = 4.01x10^{-3}x1.8x10^{-2}x5.0x10^{-2}

r = 3.61x10^{-6} M/s

7 0
2 years ago
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