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3 years ago

What process takes place when hydrogen-3 and hydrogen-2 combine to form

Chemistry

1 answer:

vazorg [7]3 years ago

4 0

Answer:

A alpha decay

hope this helps you :D

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Which of the following do scientists use to form a hypothesis?

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Explanation:

The scientific method involves making observations and asking questions. Scientists form hypotheses based on these observations and then develop controlled experiments to collecting and analyze data. Using this data, they are able to draw conclusions and form questions for new scientific research.

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2 years ago

Which of these is an example of a physical property?

Margarita [4]

The woods so far the texture is going to be A with the Henry

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2 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:

1. Convert Grams to Moles

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.

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

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

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

2. Convert Moles to Atoms

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.

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