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3 years ago

14

Which scientist was the first to conclude through experimentation that atoms contain most of their mass in a small, dense nucleu

s?

1 answer:

AURORKA [14]3 years ago

4 0

Answer:

Ernest Rutherford

Explanation:

Ernest Rutherford was the first to conclude experiments

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202H8(9) + 7 O2(g) > 4CO2(g) + 6 H2O(9).

lilavasa [31]

since the concentration of Carbon Dioxide will increase, it would make Q > K, cause equilibrium to shift in the direction with less moles of gas to alleviate the extra pressure. In this case, the reaction will shift left because there are fewer moles of gas present.

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3 years ago

Help plz:)))I’ll mark u Brainliest

Triss [41]

Answer:

Explanation:

We want the energy required for the transition:

CO 2 ( s ) + Δ → C O 2 ( g )

Explanation:

We assume that the temperature of the gas and the solid are EQUAL.

And thus we simply have to work out the product:

2 × 10^ 3 ⋅ g × 196.3 ⋅ J ⋅ g − 1 to get an answer in Joules as required.

What would be the energy change for the reverse transition:

C O 2 ( g ) + → C O 2 ( s ) ?

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3 years ago

Matches=from Chemical energy to light energy+heat energy

solong [7]

Explanation:

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3 years ago

How many atoms are in 25.00 g of B.

iren2701 [21]

Answer:

$\boxed {\boxed {\sf 1.393 *10^{24} \ atoms \ B}}$

Explanation:

<u>1. Convert Grams to Moles</u>

Use the molar mass (found on the Periodic Table) to convert from grams to moles.

Boron (B): 10.81 g/mol

Use this value as a ratio.

$\frac {10.81 \ g \ B }{1 \ mol \ B}$

Multiply by the given number of grams.

$25.00 \ g \ B *\frac {10.81 \ g \ B }{1 \ mol \ B}$

Flip the ratio so the grams of boron cancel out.

$25.00 \ g \ B *\frac {1 \ mol \ B }{10.81 \ g \ B}$

$25.00 *\frac {1 \ mol \ B }{10.81 }$

$\frac {25.00 \ mol \ B }{10.81 }=2.312673451 \ mol \ B$

<u>2. Convert Moles to Atoms</u>

We use Avogadro's Number, 6.02*10²³: the number of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case, the particles are atoms of boron.

$\frac {6.02*10^{23} \ atoms \ B} {1 \ mol \ B}$

Multiply by the number of moles we calculated.

$2.312673451 \ mol \ B *\frac {6.02*10^{23} \ atoms \ B} {1 \ mol \ B}$

The moles of boron cancel.

$2.312673451 *\frac {6.02*10^{23} \ atoms \ B} {1 }$

$2.312673451 *6.02*10^{23} \ atoms \ B} =1.39269195*10^{24} \ atoms \ B$

The original value of grams has 4 significant figures, so our answer should have the same. For the number we calculated, that is the thousandth place.

$1.392\underline69195*10^{24} \ atoms \ B$

The 6 tells us to round the 2 to a 3.

$1.393 *10^{24} \ atoms \ B$

25.00 grams of boron is equal to 1.393*10²⁴ atoms.

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3 years ago

Summarize information about sublimation

jek_recluse [69]

Sublimation is when a solid turns directly into a gas, completely bypassing the liquid phase. Dry ice is a good example.

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4 years ago