All categories

3 years ago

A(n) accepts electrons and becomes .

Chemistry

2 answers:

eimsori [14]3 years ago

6 0

Google says it's the first one.

Sever21 [200]3 years ago

6 0

Answer:

Oxidizing agent, Reduced

Explanation:

Oxidation is the loss of electrons and reduction is the gain of electrons.

An oxidizing agent oxidizes other elements but itself gets reduced. Therefore oxidizing agents accepts electrons and gets reduced.

A reducing agent reduces other elements but itself gets oxidized. Therefore reducing agents losses electrons and gets oxidized.

You might be interested in

True or false: The number of protons an element has can vary.

Lerok [7]

Answer:

true

Explanation:

5 0

2 years ago

Read 2 more answers

Does iron melt at 1728 degrees Celsius

Pepsi [2]

Iron, Wrought 1482 - 1593- 2900Iron, Gray Cast1127 - 1204 - 2200Iron, Ductile1149

7 0

2 years ago

Hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

n200080 [17]

Answer:

Explanation:

Hhhhhhhhhhhhhhhhhhhhh

5 0

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.

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