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1 year ago

Rutherford’s gold foil experiment gave evidence that an atom is mostly empty space. true false

Chemistry

1 answer:

Sholpan [36]1 year ago

4 0

The given statement is true .

<h3>What is Rutherford’s gold foil experiment?</h3>

A piece of gold foil was hit with alpha particles, which have a favorable charge. Most alpha particles went right around. This showed that the gold particles were mostly space.

The Rutherford gold leaf investigation supposed that most (99%) of all the mass of an atom is in the middle of the atom, that the nucleus is very small (105 times small than the length of the atom) and that is positively captured.

For the distribution experiment, Rutherford enjoyed a metal sheet that could be as thin as practicable. Gold is the most malleable of all known metals. It can easily be converted into very thin sheets. Hence, Rutherford established a gold foil for his alpha-ray scattering experimentation.

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The part of an experiment used to compare all the other groups

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Which element is a liquid at 1000 k <br><br> 1.Ag<br> 2.Al<br> 3.Ca<br> 4.Ni

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3 years ago

Read 2 more answers

Which acid and base react in the formation of salt sodium chloride? hydrochloric acid and sodium hydroxide sodium carbonate and

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Hydrochloric acid and sodium hydroxide

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3 years ago

glucose 6‑phosphate+H2O⟶glucose+Pi glucose 6‑phosphate+H2O⟶glucose+Pi K′eq1=270 K′eq1=270 ATP+glucose⟶ADP+glucose 6‑phosphate AT

ddd [48]

Answer:

-30.7 kj/mol

Explanation:

The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula: ∆Go ’= -RTln K’eq

where,

R = -8.315 J / mo

T = 298 K

For reaction,

1. K′eq1=270,

∆Go ’= -RTln K’eq

= - 8.315 x 298 x ln 270

= - 8.315 x 298 x 5.59

= - 13,851.293 J / mo

= - 13.85 kj/mol

2. K′eq2=890

∆Go ’= -RTln K’eq

= - 8.315 x 298 x ln 890

= - 8.315 x 298 x 6.79

= - 16.82 kj/mol

therefore, total standard free energy

= - 13.85 + (-16.82)

= -30.7 kj/mol

Thus, -30.7 kj/mol is the correct answer.

6 0

3 years ago

Write a balanced chemical equation for the standard formation reaction of liquid acetic acid hch3co2.

Afina-wow [57]

The balanced chemical equation for the standard formation reaction of liquid acetic acid is given as ,

$2C(gr) +2H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The reaction that form the products from their elements in their standard state is called formation of reaction .The acetic acid consist C , H , and O , So, determine their standard state . Carbon is graphite at 25°C and 1 atm , whereas hydrogen and oxygen are diatomic gases . Hence , we start with unbalanced reaction.

$C(gr) +H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The balanced chemical equation for the standard formation reaction of liquid acetic acid as,

$2C(gr) +2H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The combustion of liquid acetic acid is given as,

$CH_{3} COOH(l) + 2O(g)$ → $2CO_{2}((g) +2H_{2} O(l)$ ΔH =-873

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2 years ago