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3 years ago

What two types of elements make up an Ionic bond ?

1 answer:

3 0

Answer it occurs between a metal and non metal eg sodium[Na]metal and chloride[Cl]a non metal form an ionic bond to make NaCl

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Electricity is the _of charged particles. A.movement B. collection C.build up

Vitek1552 [10]

Answer:

Explanation:

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2 years ago

Which choice is not an example of a molecule? OF O H202 O 03 O NC13

MissTica

Answer:

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Explanation:

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2 years ago

Use the image to rank the colors from the color that will refract the least to the color that will refract the most.

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2 years ago

Read 2 more answers

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.

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

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.

6 0

2 years ago

When 50.0 g of 0.200 M NaCl(aq) at 24.1 °C is added to 100.0 g of 0.100 M AgNO3(aq) at 24.1 °C in a calorimeter, the temperature

Annette [7]

Answer:

Q = 693 J

Explanation:

NaCl (aq) + AgNO₃ (aq) → AgCl (s) + NaNO₃ (aq)

Q = mcΔT

Q: heat

m: mass

c: specific heat

ΔT: temperature variation

Q = (50.0g + 100.0g) x (4.20 J/g°C) x (25.2°C - 24.1°C)

Q = 693 J

4 0

3 years ago